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.
∫(ex log a + ea log a – en log x
Solution:
(ex log a + ea log a – ea log x) dx
= ∫(elog ax + elog aa – ea log xn) dx
= ∫(a1 + aa – xⁿ) dx
= [$$\frac { a^x }{log|a|}$$ + aa (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.
(ex + 1)² ex
Solution:
∫(ex + 1)² ex dx = ∫[(ex)² + 2ex + 1]ex dx
= ∫(e2x. ex + 2ex. ex + ex) dx
= ∫(e3x + 2e2x + ex) dx
= $$\frac { e^{3x} }{3}$$ + $$\frac { 2e^{2x} }{2}$$ + ex + c
= $$\frac { e^{3x} }{3}$$ + e2x + ex + 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) = ex and f(0) = 2 then find f(x)
Solution:
f'(x) = ex
Integrating both sides of the equation,
∫ f'(x) dx = ∫ ex dx
⇒ f(x) = ex + c
given f(0) = 2
2 = e0 + c
⇒ c = 1
Thus f(x) = ex + 1