Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.3

Question 1.
Integrate the following with respect to x.
∫(e^{x log a} + e^{a log a} – e^{n log x}
Solution:
(e^{x log a} + e^{a log a} – e^{a log x}) dx
= ∫(e^{log ax} + e^{log aa} – e^{a log xn}) dx
= ∫(a¹ + a^a – xⁿ) dx
= [\(\frac { a^x }{log|a|}\) + a^a (x) – \(\frac { x^{n+1} }{(n+1)}\)] + c

Question 2.
\(\frac { a^x-e^{x log b} }{e^{x log a} b^x}\)
Solution:

Question 3.
(e^x + 1)² e^x
Solution:
∫(e^x + 1)² e^x dx = ∫[(e^x)² + 2e^x + 1]e^x dx
= ∫(e^2x. e^x + 2e^x. e^x + e^x) dx
= ∫(e^3x + 2e^2x + e^x) dx
= \(\frac { e^{3x} }{3}\) + \(\frac { 2e^{2x} }{2}\) + e^x + c
= \(\frac { e^{3x} }{3}\) + e^2x + e^x + c

Question 4.
\(\frac { e^{3x}-e^{-3x} }{e^{x}}\)
Solution:

Question 5.
\(\frac { e^{3x}+e^{5x} }{e^{x}+e^{-x}}\)
Solution:

Question 6.
[1 – \(\frac { 1 }{x^2}\)] e^{[x + \(\frac { 1 }{x}\)]}
Solution:

Question 7.
\(\frac { 1 }{x(log x)^2}\)
Solution:

Question 8.
Given f'(x) = e^x and f(0) = 2 then find f(x)
Solution:
f'(x) = e^x
Integrating both sides of the equation,
∫ f'(x) dx = ∫ e^x dx
⇒ f(x) = e^x + c
given f(0) = 2
2 = e⁰ + c
⇒ c = 1
Thus f(x) = e^x + 1