Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

Using second fundamental theorem, evaluate the following:

Question 1.
\(\int_{0}^{1}\) e2x dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 2.
\(\int_{0}^{1/4}\) \(\sqrt { 1 -4x}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 2

Question 3.
\(\int_{0}^{1}\) \(\frac { xdx }{x^2+1}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 4.
\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)
Solution:
\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)
= {log |1 + ex|}\(_{0}^{3}\)
= log |1 + e³| – log |1 + e°|
= log |1 + e³| – log |1 + 1|
= log |1 + e³| – log |2|
= log |\(\frac { 1+e^3 }{2}\)|

Question 5.
\(\int_{0}^{1}\) xe dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 6.
\(\int_{1}^{e}\) \(\frac { dx }{x(1+logx)^3}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 5

Question 7.
\(\int_{-1}^{1}\) \(\frac { 2x+3 }{x^2+3x+7}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 6

Question 8.
\(\int_{0}^{π/2}\) \(\sqrt { 1 +cosx} \) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 7

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 9.
\(\int_{1}^{2}\) \(\frac { x-1 }{x^2}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 8

Evaluate the following

Question 10.
\(\int_{1}^{4}\) f(x) dx where f(x) = \(\left\{\begin{array}{l}
4 x+3,1 \leq x \leq 2 \\
3 x+5,2 \end{array}\right.\)
Solution:
\(\int_{1}^{4}\) f(x) dx
= \(\int_{1}^{2}\) f(x) dx + \(\int_{2}^{4}\) f(x) dx
= \(\int_{1}^{2}\) (4x + 3) dx + \(\int_{2}^{4}\) (3x + 5) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 9
(8 + 6) – [5] + [24 + 20] – [6 + 10]
= 14 – 5 + 44 – 16
= 58 – 21
= 37

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 11.
\(\int_{0}^{2}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}
3-2 x-x^{2}, & x \leq 1 \\
x^{2}+2 x-3, & 1<x \leq 2
\end{array}\right.\)
Solution:
\(\int_{0}^{2}\) f(x) dx
= \(\int_{0}^{1}\) f(x) dx + \(\int_{1}^{2}\) f(x) dx
= \(\int_{0}^{1}\) (3 – 2x – x²) dx + \(\int_{1}^{2}\) (x² + 2x – 3) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 10

Question 12.
\(\int_{-1}^{1}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}
x, & x \geq 0 \\
-x, & x<0
\end{array}\right.\)
Solution:
\(\int_{-1}^{1}\) f(x) dx
\(\int_{-1}^{0}\) f(x) dx + \(\int_{0}^{1}\) f(x) dx
= \(\int_{-1}^{0}\) (-x) dx + \(\int_{0}^{1}\) x dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8 11

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

Question 13.
f(x) = \(\left\{\begin{array}{l}
c x, \quad 0<x<1 \\
0, \text { otherwise }
\end{array}\right.\) find ‘c’ if \(\int_{0}^{1}\) f(x) dx = 2
Solution:
Given
f(x) = \(\left\{\begin{array}{l}
c x, \quad 0<x<1 \\
0, \text { otherwise }
\end{array}\right.\)
⇒ \(\int_{0}^{1}\) f(x) dx = 2
⇒ \(\int_{0}^{1}\) cx dx = 2
c[ \(\frac { x^2 }{ 2 }\) ]\(_{0}^{1}\) = 2
c[ \(\frac { 1 }{ 2 }\) – 0 ] = 2
\(\frac { 1 }{ 2 }\) = 2
⇒ c = 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.8

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Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.9

Evaluate the following using properties of definite integrals:

Question 1.
\(\int_{-π/4}^{π/4}\) x³ cos³ x dx
Solution:
Let f(x) = x³cos³x
f(-x) = (-x)³ cos³(-x)
= -x³ cos³x
f(-x) = -f(x)
⇒ f(x) is an odd function
∴ \(\int_{-π/4}^{π/4}\) x³ cos³ x dx = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

Question 2.
\(\int_{-π/2}^{π/2}\) sin² θ dθ
Solution:
Let f(θ)= sin² θ
f(-θ) = sin² (-θ) = [sin (-θ)]²
= [-sin θ]² = sin² θ
f(-θ) = f(θ)
∴ f(θ) is an even function
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

Question 3.
\(\int_{-1}^{1}\) log(\(\frac { 2-x }{2+x}\)) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9 2

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

Question 4.
\(\int_{0}^{π/2}\) \(\frac { sin^7x }{sin^7x+cos^7x}\) dx
Solution:
Using the property
\(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9 3

Question 5.
\(\int_{0}^{1}\) log (\(\frac { 1 }{x}\) – 1) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

Question 6.
\(\int_{0}^{1}\) \(\frac { x }{(1-x)^{3/4}}\) dx
Solution:
Let I = \(\int_{0}^{1}\) log \(\frac { x }{(1-x)^{3/4}}\) dx
Using the property
\(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9 5

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.9

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Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.2

Question 1.
The cost of an overhaul of an engine is Rs 10,000 The Operating cost per hour is at the rate of 2x-240 where the engine has run x km. find out the total cost of the engine run for 300 hours after overhaul.
Solution:
Given that the overhaul cost is Rs. 10,000.
The marginal cost is 2x – 240
MC = 2x – 240
C = ∫ MC dx + k
C = x2 – 240x + k
k is the overhaul cost
⇒ k = 10,000
So C = x2 – 240x + 10,000
When x = 300 hours, total cost is
C = (300)2 – 240(300) + 10,000
⇒ C = 90,000 – 72000 + 10,000
⇒ C = 28,000
So the total cost of the engine run for 300 hours after the overhaul is ₹ 28,000.

Question 2.
Elasticity of a function \(\frac { Ey }{Ex}\) is given by \(\frac { Ey }{Ex}\) = \(\frac { -7x }{(1-2x)(2+3x)}\). Find the function when x = 2, y = \(\frac { 3 }{8}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 1
Put x = 0
7 = A (3(0) + 2) + B (2(0) – 1)
7 = A (2) + B (-1)
7 = (2) (2) – B
B = 4 – 7
B = -3
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 2

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 3.
The Elasticity of demand with respect to price for a commodity is given by where \(\frac { (4-x)}{x}\) p is the price when demand is x. find the demand function when the price is 4 and the demand is 2. Also, find the revenue function
Solution:
The elasticity at the demand
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 3
Integrating on both sides
∫\(\frac { 1}{(x-4)}\) = ∫\(\frac { 1}{p}\) dp
log |x – 4| = log |p| + log k
log |x – 4| = log |pk| ⇒ (x – 4) = pk ……… (1)
when p = 4 and x = 2
(2 – 4) = 4k ⇒ -2 = 4k
k = -1/2
Eqn (1) ⇒ (x – 4) = p(-1/2)
-2 (x – 4) = p ⇒ p = 8 – 2x
Revenue function R = px = (8 – 2x)x
R = 8x – 2x²

Question 4.
A company receives a shipment of 500 scooters every 30 days. From experience it is known that the inventory on hand is related to the number of days x. Since the shipment, I (x) = 500 – 0.03 x², the daily holding cost per scooter is Rs 0.3. Determine the total cost for maintaining inventory for 30 days
Solution:
Here I (x) = 500 – 0.03 x²
C1 = Rs 0.3
T = 30
Total inventory carrying cost
= C1 \(\int _{0}^{T}\) I(x) dx
= 0.3 \(\int _{0}^{30}\) (500 – 0.03 x²)dx
= 0.3 [500 x – 0.03(\(\frac { x^3 }{3}\))]\( _{0}^{30}\)
= 0.3 [ 500 x – 0.01 x³]\( _{0}^{30}\)
= 0.3 [500(30) – 0.01 (30)³] – [0]
= 0.3 [15000 – 0.01 (27000)]
= 0.3 [15000 – 270] = 0.3 [14730]
= Rs 4,419

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 5.
An account fetches interest at the rate of 5% per annum compounded continuously an individual deposits Rs 1000 each year in his account. how much will be in the account after 5 years (e0.25 = 1.284)
Solution:
P = 1000
r = \(\frac { 5 }{1000}\) = 0.05
N = 5
Annuity = \(\int _{0}^{5}\) 1000 e0.05t dt
= 1000 [ \(\frac { e^{0.05t} }{0.05}\) ] \(_{0}^{5}\)
= \(\frac { 1000 }{0.05}\) [e0.05(5) – e0]
= 20000 [e0.25 – 1]
= 20000 [1.284 – 1]
= 20000 [0.284]
= Rs 5680

Question 6.
The marginal cost function of a product is given by \(\frac { dc }{dx}\) = 100 – 10x + 0.1 x² where x is the output. Obtain the total and average cost function of the firm under the assumption, that its fixed cost is t 500
Solution:
\(\frac { dc }{dx}\) = 100 – 10x + 0.1 x² and k = Rs 500
dc = (100 – 10x + 0.1 x²) dx
Integrating on both sides,
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 4

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 7.
The marginal cost function is M.C = 300 x2/5 and the fixed cost is zero. Find the total cost as a function of x
Solution:
M.C = 300 x2/5 and fixed cost K = 0
Total cos t = ∫M.C dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 5

Question 8.
If the marginal cost function of x units of output is \(\frac { a }{\sqrt {ax+b}}\) and if the cost of output is zero. Find the total cost as a function of x.
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 6
∴ C(x) = 2(ax + b)1/2 + k …….. (1)
When x = 0
eqn (1) ⇒ 0 = 2 [a(0) + b]1/2 + k
k = -2(b)1/2 ⇒ k = -2√b
Required cost function
C(x) = 2(ax + b)1/2 – 2√b
∴ C = 2\(\sqrt { ax + b}\) – 2√b

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 9.
Determine the cost of producing 200 air conditioners if the marginal cost (is per unit) is C'(x) = \(\frac { x^2 }{200}\) + 4
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 7
= 13333.33 + 800
∴ Cost of producing 200 air conditioners
= Rs 14133.33

Question 10.
The marginal revenue (in thousands of Rupees) function for a particular commodity is 5 + 3 e-0.03x where x denotes the number of units sold. Determine the total revenue from the sale of 100 units (given e-3 = approximately)
Solution:
The marginal Revenue (in thousands of Rupees) function
M.R = 5 + 3-0.03x
Total Revenue from sale of 100 units is
Total Revenue T.R = \(\int _{0}^{ 100}\) M.R dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 8
= [500 – 100 e-3] – [0 – 100 e°]
= [500 -100 (0.05)] – [-100 (1)]
= [500 – 5]+ 100
= 495 + 100 = 595 thousands
= 595 × 1000
∴ Revenue R = Rs 595000

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 11.
If the marginal revenue function for a commodity is MR = 9 – 4x². Find the demand function.
Solution:
Marginal Revenue function MR = 9 – 4x²
Revenue function R = ∫MR dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 9

Question 12.
Given marginal revenue function \(\frac { 4 }{(2x+3)^2}\) -1, show that the average revenue function is P = \(\frac { 4 }{6x+9}\) -1
Solution:
M.R = \(\frac { 4 }{(2x+3)^2}\) -1
Total Revenue R = ∫M.R dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 10
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 11

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 13.
A firms marginal revenue functions is M.R = 20 e-x/10 Find the corresponding demand function.
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 12

Question 14.
The marginal cost of production of a firm is given by C’ (x) = 5 + 0.13 x, the marginal revenue is given by R’ (x) = 18 and the fixed cost is Rs 120. Find the profit function.
Solution:
MC = C'(x) = 5 + 0.13x
C(x) = ∫C'(x) dx + k1
= ∫(5 + 0.13x) dx + k1
= 5x + \(\frac{0.13}{2} x^{2}\) + k1
When quantity produced is zero, fixed cost is 120
(i.e) When x = 0, C = 120 ⇒ k1 = 120
Cost function is 5x + 0.065x2 + 120
Now given MR = R'(x) = 18
R(x) = ∫18 dx + k2 = 18x + k2
When x = 0, R = 0 ⇒ k2 = 0
Revenue = 18x
Profit P = Total Revenue – Total cost = 18x – (5x + 0.065x2 + 120)
Profit function = 13x – 0.065x2 – 120

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 15.
If the marginal revenue function is R'(x) = 1500 – 4x – 3x². Find the revenue function and average revenue function.
Solution:
Given marginal revenue function
MR = R’(x)= 1500 – 4x – 3x2
Revenue function R(x) = ∫R'(x) dx + c
R = ∫(1500 – 4x – 3x2) dx + c
R = 1500x – 2x2 – x3 + c
When x = 0, R = 0 ⇒ c = 0
So R = 1500x – 2x2 – x3
Average revenue function P = \(\frac{R}{x}\) ⇒ 1500 – 2x – x2

Question 16.
Find the revenue function and the demand function if the marginal revenue for x units MR = 10 + 3x – x
Solution:
The marginal revenue function
MR = 10 + 3x – x²
The Revenue function
R = ∫(MR) dx
= ∫(10 + 3x – x²)dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 13

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 17.
The marginal cost function of a commodity is given by Mc = \(\frac { 14000 }{\sqrt{7x+4}}\) and the fixed cost is Rs 18,000. Find the total cost average cost.
Solution:
The marginal cost function of a commodity
Mc = \(\frac { 14000 }{\sqrt{7x+4}}\) = 14000 (7x + 4)-1/2
Fixed cost k = Rs 18,000
Total cost function C = ∫(M.C) dx
= ∫14000 (7x + 4)-1/2 dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 14

Question 18.
If the marginal cost (MC) of production of the company is directly proportional to the number of units (x) produced, then find the total cost function, when the fixed cost is Rs 5,000 and the cost of producing 50 units is Rs 5,625.
Solution:
M.C αx
M.C = λx
fixed cost k = Rs 5000
Cost function C = ∫(M.C) dx
= ∫λx dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2 15

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

Question 19.
If MR = 20 – 5x + 3x², Find total revenue function
Solution:
MR = 20 – 5x + 3x²
Total Revenue function
R = ∫(MR) dx = ∫(20 – 5x + 3x²) dx
R = 20x – \(\frac { 5x^2 }{2}\) + \(\frac {3x^3 }{3}\) + k
when x = 0; R = 0 ⇒ k = 0
∴ R = 20x – \(\frac { 5 }{2}\) x² + x³

Question 20.
If MR = 14 – 6x + 9x², Find the demand function.
Solution:
MR = 14 – 6x + 9x2
R = ∫(14 – 6x + 9x2) dx + k
= 14x – 3x2 + 3x3 + k
Since R = 0, when x = 0, k = 0
So revenue function R = 14x – 3x2 + 3x3
Demand function P = \(\frac{R}{x}\) = 14 – 3x + 3x2

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.2

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Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.1

Question 1.
Using Interation, find the area of the region bounded the line given is 2y + x = 8, the x axis and the lines x = 2, x = 4.
Solution:
The equation of the line given is 2y + x = 8
⇒ 2y = 8 – x ⇒ y = \(\frac { 8-x }{2}\)
y = 4 – \(\frac { x }{2}\)
Also x varies from 2 to 4
The required Area
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 1
= (16 – 4) – (8 – 1)
= 12 – 7 = 5 sq.units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Question 2.
Find the area bounded by the lines y – 2x – 4 = 0, y = 0, y = 3 and the y-axis.
Solution:
The equation of the line given is y – 2x – 4 = 0
⇒ 2x = y – 4 ⇒ x = \(\frac { y-4 }{2}\)
∴ x = \(\frac { y }{2}\) – 2
Also y varies from 1 to 3
Required Area
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 2
= 2 – 4 = -2
Area can’t be in negative.
∴ Area = 2 sq.units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Question 3.
Calculate the area bounded by the parabola y² = 4ax and its latus rectum.
Solution:
Given parabola is y2 = 4ax
Its focus is (a, 0)
Equation of latus rectum is x = a
The parabola is symmetrical about the x-axis
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 3

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Question 4.
Find the area bounded by the line y = x and x-axis and the ordinates x = 1, x = 2
Solution:
The equation of the given line is y = x and x varies from 1 to 2
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 4

Question 5.
Using integration, find the area of the region bounded by the line y – 1 = x, the x-axis and the ordinates x = -2, x = 3.
Solution :
The equation of given line is y – 1 = x
y = x + 1
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 5
The line y = x + 1 meets the x-axis at x = -1
Since x varies from -2 to 3
Hence a part of lies below the x-axis and the other part lies above the x-axis.
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 6

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Question 6.
Find the area of the region lying in the first quadrant bounded by the region y = 4x², x = 0, y = 0 and y = 4
Solution:
The equation of a parabola given is y = 4x²
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 7
Area of the region lying in the first quadrant
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 8

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

Question 7.
Find the area bounded by the curve y = x² and the line y = 4
Solution:
Equation of the curve y = x² ………. (1)
Equation of the line y = 4 ………. (2)
Solving equation (1) & (2)
x² = 4 ⇒ x = ± 2
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1 9

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.1

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Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.12 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.12

Choose the most suitable answer from the given four alternatives:

Question 1.
∫\(\frac { 1 }{x^3}\) dx is
(a) \(\frac { -3 }{x^2}\) + c
(b) \(\frac { -1 }{2x^2}\) + c
(c) \(\frac { -1 }{3x^2}\) + c
(d) \(\frac { -2 }{x^2}\) + c
Solution:
(b) \(\frac { -1 }{2x^2}\) + c
Hint:
∫\(\frac { 1 }{x^3}\) dx = ∫x-3 dx = [ \(\frac { x^{-3+1} }{-3+1}\) ] + c
= (\(\frac { x^{-2} }{-2}\)) + c = \(\frac { -1 }{2x^2}\) + c

Question 2.
∫2x dx is
(a) 2x log 2 + c
(b) 2x + c
(c) \(\frac { 2^x }{log 2}\) + c
(d) \(\frac { log 2 }{2^x}\) + c
Solution:
(c) \(\frac { 2^x }{log 2}\) + c
Hint:
∫2x dx = ∫ax dx = \(\frac { a^x }{log a}\) + c

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 3.
∫\(\frac { sin 2x }{2 sin x}\) dx is
(a) sin x + c
(b) \(\frac { 1 }{2}\) sin x + c
(c) cos x + c
(d) \(\frac { 1 }{2}\) cos x + c
Solution:
(a) sin x + c
Hint:
∫\(\frac { sin 2x }{2 sin x}\) dx = ∫\(\frac { 2sin x cos x }{2 sin x}\) dx
= ∫cos x dx
= sin x + c

Question 4.
∫\(\frac { sin 5x-sin x }{cos 3x}\) dx is
(a) -cos 2x + c
(b) -cos 2x – c
(c) –\(\frac { 1 }{4}\) cos 2x + c
(d) -4 cos 2x + c
Solution:
(a) -cos 2x + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 5.
∫\(\frac { log x}{x}\) dx, x > 0 is
(a) \(\frac { 1 }{2}\) (log x)² + c
(b) –\(\frac { 1 }{2}\) (log x)²
(c) \(\frac { 2 }{x^2}\) + c
(d) \(\frac { 2 }{x^2}\) – c
Solution:
(a) \(\frac { 1 }{2}\) (log x)² + c
Hint:
∫\(\frac { log x}{x}\) dx, x > 0
∫ tdt = [ \(\frac { t^2 }{2}\) ] + c
= \(\frac { (log x)^2 }{2}\) + c
let t = log x
\(\frac { dt }{dx}\) = \(\frac { 1 }{x}\)
dt = \(\frac { 1 }{x}\) dx

Question 6.
∫\(\frac { e^x }{\sqrt{1+e^x}}\) dx is
(a) \(\frac { e^x }{\sqrt{1+e^x}}\) + c
(b) 2\(\sqrt{1+e^x}\) + c
(c) \(\sqrt{1+e^x}\) + c
(d) ex\(\sqrt{1+e^x}\) + c
Solution:
(b) 2\(\sqrt{1+e^x}\) + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 2

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 7.
∫\(\sqrt { e^x}\) dx is
(a) \(\sqrt { e^x}\) + c
(b) 2\(\sqrt { e^x}\) + c
(c) \(\frac { 1 }{2}\) \(\sqrt { e^x}\) + c
(d) \(\frac { 1 }{2\sqrt { e^x}}\) + c
Solution:
(b) 2\(\sqrt { e^x}\) + c
Hint:
∫\(\sqrt { e^x}\) dx
= ∫\(\sqrt { e^x}\) dx = ∫(ex)1/2 dx = ∫ ex/2 dx
= \(\frac { e^{x/2} }{1/2}\) + c = 2ex/2 + c
= 2(ex)1/2 + c = 2\(\sqrt { e^x}\) + c

Question 8.
∫e2x [2x² + 2x] dx
(a) e2x x² + c
(b) xe2x + c
(c) 2x²e² + c
(d) \(\frac { x^2e^x }{2}\) + c
Solution:
(a) e2x x² + c
Hint:
∫e2x (2x² + 2x) dx
Let f(x) = x²; f'(x) = 2x and a = 2
= ∫eax [af(x),+ f ’(x)] = eax f(x) + c
= ∫e2x (2x² + 2x) dx = e2x (x²) + c

Question 9.
\(\frac { e^x }{e^x+1}\) dx is
(a) log |\(\frac { e^x }{e^x+1}\)| + c
(b) log |\(\frac { e^x+1 }{e^x}\)| + c
(c) log |ex| + c
(d) log |ex + 1| + c
Solution:
(d) log |ex + 1| + c
Hint:
∫\(\frac { e^x }{e^x+1}\) dx
= ∫\(\frac { dt }{t}\)
= log |t| + c
= log |ex + 1| + c
take t = ex + 1
\(\frac { dt }{dx}\) = ex
dt = ex dx

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 10.
∫\(\frac { 9 }{x-3}-\frac { 1 }{x+1}\) dx is
(a) log |x – 3| – log|x + 1| + c
(b) log|x – 3| + log|x + 1| + c
(c) 9 log |x – 3| – log |x + 1| + c
(d) 9 log |x – 3| + log |x + 1| + c
Solution:
(c) 9 log |x – 3| – log |x + 1| + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 3

Question 11.
∫\(\frac { 2x^3 }{4+x^4}\) dx is
(a) log |4 + x4| + c
(b) \(\frac { 1 }{2}\) log |4 + x4| + c
(c) \(\frac { 1 }{2}\) log |4 + x4| + c
(d) log |\(\frac { 2x^3 }{4+x^4}\) + c
Solution:
(b) \(\frac { 1 }{2}\) log |4 + x4| + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 4

Question 12.
∫\(\frac { dx }{\sqrt{x^2-36}}\) is
(a) \(\sqrt{x^2-36}\) + c
(b) log |x + \(\sqrt{x^2-36}\)| + c
(c) log |x – \(\sqrt{x^2-36}\)| + c
(d) log |x² + \(\sqrt{x^2-36}\)| + c
Solution:
(b) log |x + \(\sqrt{x^2-36}\)| + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 5

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 13.
∫\(\frac { 2x+3 }{\sqrt{x^2+3x+2}}\) dx is
(a) \(\sqrt{x^2+3x+2}\) + c
(b) 2\(\sqrt{x^2+3x+2}\) + c
(c) \(\sqrt{x^2+3x+2}\) + c
(d) \(\frac { 2 }{3}\) (x² + 3x + 2) + c
Solution:
(b) 2\(\sqrt{x^2+3x+2}\) + c
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 6

Question 14.
\(\int_{0}^{4}\) (2x + 1) dx is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
\(\int_{0}^{4}\) (2x + 1) dx
= [2(\(\frac { x^2 }{2}\)) + x]\(_{0}^{1}\) = [x² + x]\(_{0}^{1}\)
= [(1)² + (1)] – [0] = 2

Question 15.
\(\int_{2}^{4}\) \(\frac { dx }{x}\) is
(a) log 4
(b) 0
(c) log 2
(d) log 8
Solution:
(c) log 2
Hint:
\(\int_{2}^{4}\) \(\frac { dx }{x}\)
\(\int_{2}^{4}\) \(\frac { dx }{x}\) = [log |x|]\(_{0}^{1}\) = log |4| – log |2|
= log[ \(\frac { 4}{2}\) ] = log 2

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 16.
\(\int_{0}^{∞}\) e-2x dx is
(a) 0
(b) 1
(c) 2
(d) \(\frac { 1 }{2}\)
Solution:
(d) \(\frac { 1 }{2}\)
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 7

Question 17.
\(\int_{-1}^{1}\) x³ ex4 dx is
(a) 1
(b) 2\(\int_{0}^{1}\) x³ ex4
(c) 0
(d) ex4
Solution:
(c) 0
Hint:
\(\int_{-1}^{1}\) x³ ex4 dx
Let f (x) = x³ex4
f(-x) = (-x)² e(-x)4
= -x² ex4
f(-x) = -f(x)
⇒ f(x) is an odd function
∴ \(\int_{-1}^{1}\) x³ ex4 dx = 0

Question 18.
If f(x) is a continuous function and a < c < b, then \(\int_{a}^{c}\) f(x) dx + \(\int_{c}^{b}\) f(x) dx is
(a) \(\int_{a}^{b}\) f(x) dx – \(\int_{a}^{c}\) f(x) dx
(b) \(\int_{a}^{c}\) f(x) dx – \(\int_{a}^{b}\) f(x) dx
(c) \(\int_{a}^{b}\) f(x) dx
(d) 0
Solution:
(c) \(\int_{a}^{b}\) f(x) dx

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 19.
The value of \(\int_{-π/2}^{π/2}\) cos x dx is
(a) 0
(b) 2
(c) 1
(d) 4
Solution:
(b) 2
Hint:
\(\int_{-π/2}^{π/2}\) cos x dx
Let f(x) = cos x
f(-x) = cos (-x) = cos (x) = f(x)
∴ f(x) is an even function
\(\int_{-π/2}^{π/2}\) cos x dx = 2 × \(\int_{0}^{π/2}\) cos x dx
= 2 × [sin x]\(_{0}^{-π/2}\) = 2 [sin π/2 – sin 0]
= 2 [1 – 0] = 2

Question 20.
\(\int_{-π/2}^{π/2}\) \(\sqrt {x^4(1-x)^2}\) dx
(a) \(\frac { 1 }{12}\)
(b) \(\frac { -7 }{12}\)
(c) \(\frac { 7 }{12}\)
(d) \(\frac { -1 }{12}\)
Solution:
(a) \(\frac { 1 }{12}\)
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 8

Question 21.
If \(\int_{0}^{1}\) f(x) dx = 1, \(\int_{0}^{1}\) x f(x) dx = a and \(\int_{0}^{1}\) x² f(x) dx = a², then \(\int_{0}^{1}\) (a – x)² f(x) dx is
(a) 4a²
(b) 0
(c) a²
(d) 1
Solution:
(b) 0
Hint:
\(\int_{0}^{1}\) (a – x)² f(x) dx
= \(\int_{0}^{1}\) [a² +x² – 2ax] f(x) dx
= \(\int_{0}^{1}\) a² + f (x) dx + \(\int_{0}^{1}\) x² f (x) dx – 2a\(\int_{0}^{1}\) x f(x) dx
= a²(1) + a² – 2a(a) – 2a² – 2a² = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 22.
The value of \(\int_{2}^{3}\) f(5 – x) dx – \(\int_{2}^{3}\) f(x) dx is
(a) 1
(b) 0
(c) -1
(d) 5
Solution:
(b) 0
Hint:
\(\int_{2}^{3}\) f(5 – x) dx – \(\int_{2}^{3}\) f(x) dx
Using the property
= \(\int_{2}^{3}\) f(x) dx = \(\int_{a}^{b}\) f(a + b – x) dx
= \(\int_{2}^{3}\) f (5 – x) – \(\int_{2}^{3}\) f (5 – x) dx
= 0

Question 23.
\(\int_{0}^{4}\) (√x + \(\frac { 1 }{√x}\)), dx is
(a) \(\frac { 20 }{3}\)
(b) \(\frac { 21 }{3}\)
(c) \(\frac { 28 }{3}\)
(d) \(\frac { 1 }{3}\)
Solution:
(c) \(\frac { 28 }{3}\)
Hint:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12 9

Question 24.
\(\int_{0}^{π/3}\) tan x dx is
(a) log 2
(b) 0
(c) log √2
(d) 2 log 2
Solution:
(a) log 2
Hint:
\(\int_{0}^{π/3}\) tan x dx
= ∫tan x dx
= ∫\(\frac { sin x }{cos x}\) dx
= -∫\(\frac { -sin x }{cos x}\) dx
= -log |cos x| + c
= log sec x + c
= [log (sec x)]\(_{0}^{π/3}\)
= log [(sec π/3) – log (sec 0)]
= log (2) – log (1)
= log 2 – (0) = log 2

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 25.
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8
(a) 5040
(b) 5400
(c) 4500
(d) 5540
Solution:
(a) 5040
Hint:
\(\Upsilon\) (8) = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Question 26.
Γ(n) is
(a) (n – 1)!
(b) n!
(c) n Γ (n)
(d) (n – 1) Γ(n)
Solution:
(a) (n – 1)!
Hint:
Γ(n) = Γ(n – 1) + 1 = (n – 1)!

Question 27.
Γ(1) is
(a) 0
(b) 1
(c) n
(d) n!
Solution:
(b) 1
Hint:
\(\Upsilon\) (1) = 0! = 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

Question 28.
If n > 0, then Γ(n) is
(a) \(\int_{0}^{1}\) e-x xn-1 dx
(b) \(\int_{0}^{1}\) e-x xⁿ dx
(c) \(\int_{0}^{∞}\) ex x-n dx
(d) \(\int_{0}^{∞}\) e-x xn-1 dx
Solution:
(d) \(\int_{0}^{∞}\) e-x xn-1 dx

Question 29.
Γ(\(\frac { 3 }{2}\))
(a) √π
(b) \(\frac { √π }{2}\)
(c) 2√π
(d) \(\frac { 3 }{2}\)
Solution:
(b) \(\frac { √π }{2}\)
Hint:
\(\Upsilon\) (3/2) = \(\frac { 2 }{2}\) \(\Upsilon\) [ \(\frac { 3 }{2}\) ]
= \(\frac { 3 }{2}\) √π

Question 30.
\(\int_{0}^{∞}\) x4 e-x dx is
(a) 12
(b) 4
(c) 4!
(d) 64
Solution:
(b) \(\frac { √π }{2}\)
Hint:
\(\int_{0}^{∞}\) x4 e-x dx
= ∫xⁿ e-ax dx = \(\frac { n! }{a{n+1}}\)
= \(\frac { 4! }{(1)^{n+1}}\)
= \(\frac { 4! }{(1)^5}\)
= 4!

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.12

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Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Miscellaneous Problems

Question 1.
∫\(\frac { 1 }{\sqrt{x-1}-\sqrt{x+3}}\) dx
Solution:
∫\(\frac { 1 }{\sqrt{x+2}-\sqrt{x+3}}\) dx
Conjugating the Denominator
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Question 2.
∫\(\frac { dx }{2-3x-2x^2}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 2

Question 3.
∫\(\frac { dx }{e^x+6+5e^{-x}}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Question 4.
∫\(\sqrt { 2x^2-3}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 4

Question 5.
∫\(\sqrt { 9x^2+12x+3}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 5
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 6

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Question 6.
∫(x + 1)² log x dx
Solution:
∫udv = uv – ∫vdu
∫(x + 1)² log x dx
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 7

Question 7.
∫log (x – \(\sqrt { x^2-1}\)) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 8
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 9

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Question 8.
\(\int_{0}^{1}\) \(\sqrt { x(x-1)}\) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 10

Question 9.
\(\int_{-1}^{1}\) x² e-2x dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 11

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

Question 10.
\(\int_{0}^{3}\) \(\frac { xdx }{\sqrt {x+1} + \sqrt{5x+1}}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems 12

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Miscellaneous Problems

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Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.10

Evaluate the following:

Question 1.
(i) \(\Upsilon\) (4)
Solution:
Γ(4) = Γ(3 + 1) = 3! = 6

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10

(ii) \(\Upsilon\) (\(\frac { 9 }{2}\))
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10 1

(iii) \(\int_{0}^{∞}\) e-mx x6 dx
Solution:
W.K.T \(\int_{0}^{∞}\) xⁿ e-ax dx = \(\frac { n! }{a^{n+1}}\)
∴ \(\int_{0}^{∞}\) e-mx x6 dx = \(\frac { 6! }{3^{6+1}}\) = \(\frac { 6! }{m^7}\)

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10

(iv) \(\int_{0}^{∞}\) e-4x x4 dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10 2

(v) \(\int_{0}^{∞}\) e-x/2 x5 dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10

Question 2.
If f(x) = \(\left\{\begin{array}{l}
x^{2} e^{-2 x}, x \geq 0 \\
0, \text { otherwise }
\end{array}\right.\), then evaluate \(\int_{0}^{∞}\) f(x) dx
Solution:
Given
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.10

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Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.11

Evaluate the following integrals as the limit of the sum:

Question 1.
\(\int_{0}^{1}\) (x + 4) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11

Question 2.
\(\int_{1}^{3}\) x dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11 2

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11

Question 3.
\(\int_{1}^{3}\) (2x + 3) dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11

Question 4.
\(\int_{0}^{1}\) x² dx
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.11

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Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Miscellaneous Problems

Question 1.
A manufacture’s marginal revenue functional is given by MR = 275 – x – 0.3x². Find the increase in the manufactures total revenue if the production increased from 10 to 20 units.
Solution:
MR = 275 – x – 0.3x²
The increase in the manufactures total revenue 20
T.R = ∫MR dx = (275 – x – 0.3x²) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 1
= [5500 – 200 – 0.1 (8000)] – [2750 – 50 – 0.1(1000)]
= [5500 – 200 – 800] – [2750 – 50 – 100]
= 4500 – 2600
= Rs 1900

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Question 2.
A company has determined that marginal cost function for product of a particular commodity is given by MC = 125 + 10x – \(\frac { 8 }{3}\). Where C is the cost of producing x units of the commodity. If the fixed cost Rs 250 what is cost of producing 15 units.
Solution:
MC = 125 + 10x – \(\frac { x^2 }{9}\)
Fixed cos t K = Rs 250
C = ∫MC dx – ∫(125 + 10x – \(\frac { x^2 }{9}\)) dx
C = 125x + \(\frac { 10x^2 }{9}\) – \(\frac { x^3 }{9×3}\) + k
C = 125x + 5x² – \(\frac { x^3 }{27}\) + 250
when x = 15
C = 125(15) + 5(15)² – \(\frac { (15)^3 }{27}\) + 250
= 1875 + 1125 – 125 + 250
C = Rs 3,125

Question 3.
The marginal revenue function for a firm given by MR = \(\frac { 2 }{x+3}\) – \(\frac { 2x }{(x+3)^2}\) + 5. Show that the demand function is P = \(\frac { 2x }{(x+3)^2}\) + 5
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 1
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 2

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Question 4.
For the marginal revenue function MR = 6 – 3x² – x³, Find the revenue function and demand function
Solution:
MR = 6 – 3x² – x³
Revenue function R = ∫MR dx
R = ∫(6 – 3x² – x³)dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 3

Question 5.
The marginal cost of production of a firm is given by C'(x) = 20 + \(\frac { x }{20}\) the marginal revenue is given by R’(x) = 30 and the fixed cost is Rs 100. Find the profit function.
Solution:
C'(x) = 20 + \(\frac { x }{20}\)
Fixed cost k1 = Rs 100
C(x) = ∫C1(x) dx + k1
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 4

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Question 6.
The demand equation for a product is Pd = 20 – 5x and the supply equation is Ps = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium.
Solution:
Pd = 20 – 5x and Ps = 4x + 8
At market equilibrium
Pd = Pd
20 – 5x = 4x + 8 ⇒ 20 – 8 = 4x + 5x
9x = 12 ⇒ x = \(\frac { 12 }{9}\)
∴ x = \(\frac { 4 }{3}\)
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 5
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 6

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Question 7.
A company requires f(x) number of hours to produce 500 units. lt is represented by f(x) = 1800x-0.4. Find out the number of hours required to produce additional 400 units. [(900)0.6 = 59.22, (500)0.6 = 41.63]
Solution:
f(x) number of hours to produce 500
f(x) = 1800 x-0.4
The number of hours required to produce additions
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 7

Question 8.
The price elasticity of demand for a commodity is \(\frac { p }{x^3}\). Find the demand function if the quantity of demand is 3, When the price is Rs 2.
Solution:
Price elasticity of demand
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 8

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

Question 9.
Find the area of the region bounded by the curve between the parabola y = 8x² – 4x + 6 the y-axis and the ordinate at x = 2.
Solution:
Equation of the parabola
y = 8x² – 4x + 6
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 9
The required region is bounded by the y-axis and the ordinate at x = 2.
∴ Required Area A = \(\int_{0}^{2}\)ydx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 10

Question 10.
Find the area of the region bounded by the curve y² = 27x³ and the line x = 0, y = 1 and y = 2
Solution:
Equation of the curve is y² = 27x³
⇒ x³ = \(\frac { y^2 }{27}\) = \(\frac { y^3 }{3^3}\)
∴ x = \(\frac { (y)^{2/3} }{3}\)
Since the Area of the region bounded by the given curve and the lines x = 0, y = 1 and y = 2
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems 11

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Miscellaneous Problems

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Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.3

Question 1.
Calculate consumer’s surplus if the demand function p = 50 – 2x and x = 20
Solution:
Demand function p = 50 – 2x and x = 20
when x = 20, p = 50 – 2(20)
p = 50 – 40 = 10
∴ p0 = 10
CS = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)
= \(\int _{0}^{20}\) (50 – 2x)dx – (10 × 20)
= [50x – 2(\(\frac { x^2 }{2}\))]\( _{0}^{20}\) – 200
= [50x – x²]\( _{0}^{20}\) – 200
= {50(20) – (20)² – [0]} – 200
= (1000 – 400) – 200
= 600 – 200
∴ C.S = 400 units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 2.
Calculate consumer’s surplus if the demand function p = 122 – 5x – 2x² and x = 6.
Solution:
Demand function p = 122 – 5x – 2x² and x = 6
when x = 6; p = 122 – 5(6) – 2(6)²
= 122 – 30 – 2 (36)
= 122 – 102 = 20
∴ p0 = 20
C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)
= \(\int _{0}^{6}\)(122 – 5x – 2x²) dx – (20 × 6)
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 1
[732 – 5(18) – 2(72)] – 120
= 732 – 90 – 144 – 120
= 732 – 354 = 378
∴ CS = 378 units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 3.
The demand function p = 85 – 5x and supply function p = 3x – 3. Calculate the equilibrium price and quantity demanded. Also calculate consumer’s surplus.
Solution:
Demand function p = 85 – 5x
Supply function p = 3x – 35
W.K.T. at equilibrium prices pd = ps
85 – 5x = 3x – 35
85 + 35 = 3x + 5x
120 = 8x ⇒ x = \(\frac { 120 }{8}\)
∴ x = 15
when x = 15 p0 = 85 – 5(15) = 85 – 75 = 10
C.S = \(\int _{0}^{x}\) f(x) dx – x0p0
= \(\int _{0}^{x}\) (85 – 5x) dx – (15)(10)
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 2
= 1275 – \(\frac { 1125 }{2}\) – 150
= 1275 – 562.50 – 150
= 1275 – 712.50
∴ CS = 562.50 units

Question 4.
The demand function for a commodity is p = e-x. Find the consumer’s surplus when p = 0.5
Solution:
The demand function p = e-x
when p = 0.5 ⇒ 0.5 = e-x
\(\frac { 1 }{2}\) = \(\frac { 1 }{e^x}\) ⇒ ex = 2
∴ x = log 2
∴ Consumer’s surplus
C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 3

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 5.
Calculate the producer’s surplus at x = 5 for the supply function p = 7 + x
Solution:
The supply function p = 7 + x
when x = 5 ⇒ p = 7 + 5 = 12
∴ x0 = 5 and p0 = 12
Producer’s surplus
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 4

Question 6.
If the supply function for a product is p = 3x + 5x². Find the producer’s surplus when x = 4
Solution:
The supply function p = 3x + 5x²
when x = 4 ⇒ p = 3(4) + 5(4)²
p = 12 + 5(16)= 12 + 80
p = 92
∴ x0 = 4 and p0 = 92
Producer’s Surplus
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 5

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 7.
The demand function for a commodity is p = \(\frac { 36 }{x+4}\) Find the producer’s surplus when the prevailing market price is Rs 6.
Solution:
The demand function for a commodity
p = \(\frac { 36 }{x+4}\)
when p = 6 ⇒ 6 = \(\frac { 36 }{x+4}\)
x + 4 = \(\frac { 36 }{6}\) ⇒ x + 4 = 6
x = 2
∴ p0 = 6 and x0 = 2
The consumer’s surplus
C.S = \(\int _{0}^{x}\) f(x) dx – x0p0
= \(\int _{0}^{2}\) (\(\frac { 36 }{x+4}\)) dx – 2(6)
= 36 [log (x + 4)]\(_{0}^{2}\) – 12
= 36 [log (2 + 4) – log (0 + 4)] – 12
= 36 [log6 – log4] – 12
= 36[log(\(\frac { 6 }{4}\))] – 12
∴ CS = 36 log(\(\frac { 6 }{4}\)) – 12 units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 8.
The demand and supply functions under perfect competition are pd = 1600 – x² and ps = 2x² + 400 respectively, find the producer’s surplus.
Solution:
pd = 1600 – x² and ps = 2x² + 400
Under the perfect competition pd = ps
1600 – x² = 2x² + 400
1600 – 400 = 2x² + x² ⇒ 1200 = 3x²
⇒ x² – 400 ⇒ x = 20 or -20
The value of x cannot be negative, x = 20 when x0 = 20;
p0 = 1600 – (20)² = 1600 – 400
P0 = 1200
PS = x0p0 – \(\int _{0}^{x_0}\) g(x) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 6

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 9.
Under perfect competition for a commodity the demand and supply laws are Pd = \(\frac { 8 }{x+1}\) – 2 and Ps = \(\frac { x+3 }{2}\) respectively. Find the consumer’s and producer’s surplus.
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 7
16 – (x² + 3x + x + 3) = 2 [2(x + 1)]
16 – (x² + 4x + 3) = 4(x + 1)
16 – x² – 4x – 3 = 4x + 4
x² + 4x + 4x + 4 + 3 – 16 = 0
x² + 8x – 9 = 0
(x + 9) (x – 1) = 0 ⇒ x = -9 (or) x = 1
The value of x cannot be negative x = 1 when x0 = 1
p0 = \(\frac { 8 }{1+1}\) – 2 ⇒ p0 = \(\frac { 8 }{2}\) – 2
p0 = 4 – 2 ⇒ p0 = 2
CS = \(\int _{0}^{x}\) f(x) dx – x0p0
= \(\int _{0}^{1}\) (\(\frac { 8 }{x+1}\) – 2) dx – (1) (2)
= {8[log(x + 1)] – 2x}\(\int _{0}^{1}\) – 2
= 8 {[log (1 + 1) – 2(1)] – 8 [log (0 + 1) – 2(0)]} – 2
= [8 log (2) – 2 – 8 log1] – 2
= 8 log(\(\frac { 8 }{2}\)) – 2 – 2
C.S = (8 log 2 – 4) units
P.S = x0p0 – \(\int _{0}^{x_0}\) g(x) dx
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 8

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 10.
The demand equation for a products is x = \(\sqrt {100-p}\) and the supply equation is x = \(\frac {p}{2}\) – 10. Determine the consumer’s surplus and producer’s, under market equilibrium.
Solution:
pd = \(\sqrt {100-p}\) and ps = \(\sqrt {100-p}\)
Under market equilibrium, pd = ps
\(\sqrt {100-p}\) = \(\frac {p}{2}\) – 10
Squaring on both sides
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 9
36p = p² ⇒ p² – 36 p = 0
p (p – 36) = 0 ⇒ p = 0 or p = 36
The value of p cannot be zero, ∴p0 = 36 when p0 = 36
x0 = \(\sqrt {100-36}\) = \(\sqrt {64}\)
∴ x0 = 8
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 10
= 288 – [x² + 20x]\( _{0}^{8}\)
= 288 – { [(8)² + 20(8)] – [0]}
= 288 – [64 + 160]
= 288 – 224 = 64
PS = 64 Units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

Question 11.
Find the consumer’s surplus and producer’s surplus for the demand function pd = 25 – 3x and supply function ps = 5 + 2x
Solution:
pd = 25 – 3x and ps = 5 + 2x
Under market equilibrium, pd = ps
25 – 3x = 5 + 2x
25 – 5 = 2x + 3x ⇒ 5x = 20
∴ x = 4
when x = 4
P0 = 25 – 3(4)
= 25 – 12 = 13
p0 = 13
Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3 11
= 52 – [5x + x²]\( _{0}^{4}\)
= 52 – (5(4) + (4)²) – (0)}
= 52 – [20 + 16]
= 52 – 36
∴ PS = 16 units

Samacheer Kalvi 12th Business Maths Guide Chapter 3 Integral Calculus II Ex 3.3

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