# Integral Calculus – CA Foundation Maths Study Material

This Integral Calculus – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Integral Calculus – CA Foundation Maths Study Material

Previous Year Exam Questions

Question 1.
$$\int_0^1$$(ex + e-x) dx is:
(a) e – e-1
(b) e-1 – e
(c) e + e1
(d) None
Solution :
(a) is correct = e – 1 – e-1 + 1 = e – e-1

Question 2.
∫$$\frac{8 x^2}{\left(x^3+2\right)^3}$$dx is equal to: [1 Mark, Nov. 2006]
(a) –$$\frac{4}{3}$$(x3 + 2) + C
(b) –$$\frac{4}{3}$$(x3 + 2)-2 + C
(c) $$\frac{4}{3}$$(x3 + 2)2 + C
(d) None of these  Question 3.
∫$$\frac{d x}{\sqrt{x^2+a^2}}$$:
(a) $$\frac{1}{2}$$log(x + $$\sqrt{x^2+a^2}$$) + C
(b) log(x + $$\sqrt{x^2+a^2}$$) + C
(c) log(x$$\sqrt{x^2+a^2}$$) + C
(d) $$\frac{1}{2}$$log(x$$\sqrt{x^2+a^2}$$) + C
(b) is correct
I. Remember it as Formula
Trick II. Go by choices
Differentiation of which option is the Integration of given Function
Here to differentiate is more easy than to Integrate. III Detail Method
Let t – x = $$\sqrt{x^2+a^2}$$ Question 4.
The value of $$\int_0^2 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}$$dx is: [1 Mark, Feb. 2007 & May 2007]
(a) 0
(b) 3
(c) 2
(d) 1
(d) is correct 2I = 2; I = 1

Question 5.
The Integral of (e3x + e-3x)/ex is: [1 Mark, May 2007]
(a) $$\frac{e^{2 x}}{2}+\frac{e^{-4 x}}{4}$$ + C
(b) $$\frac{e^{2 x}}{2}-\frac{e^{-4 x}}{4}$$ + C
(c) e2x – e-4x
(d) None of these
(b) is correct Where c = Integration (Arbitrary) Constant. Question 6.
∫x2e3xdx is:
(a) x2.e3x – 2xe3x + 2e3x + C
(b) $$\frac{e^{3 x}}{3}-\frac{x \cdot e^{3 x}}{9}$$ + 2e3x + C
(c) $$\frac{x^2 \cdot e^{3 x}}{3}-\frac{2 x \cdot e^{3 x}}{9}+\frac{2}{27}$$e3x + C
(d) None of these
(c) is correct.
By parts; we get Question 7.
$$\int_1^2 \frac{2 x}{1+x^2}$$dx: [1 Mark, May & Aug. 2007]
(a) loge$$\frac{5}{2}$$
(b) loge5 – loge2 + 1
(a) loge$$\frac{2}{5}$$
(d) None of these
(a) = log(1+22) – log(1 + 12)
= log 5 – log 2 = log$$\frac{5}{2}$$
(a) is correct

Question 8.
The value of $$\int_1^e \frac{(1+\log x)}{x}$$dx is: [Given Loge = 1]. [1 Mark, Aug. 2007]
(a) 1/2
(b) 3/2
(c) 1
(d) 5/2
(b) is correct Question 9.
Find ∫$$\frac{x^3}{\left(x^2+1\right)^3}$$dx: [1 Mark, Aug. 2007] (b) is correct Question 10.
∫$$\frac{1}{x^2-a^2}$$dx is: [1 Mark, Nov. 2007]
(a) log(x – a) – log(x + a) + C
(b) log x – $$\frac{a}{x+a}$$ + C
(c) $$\frac{1}{2 a}$$ log$$\left(\frac{x-a}{x+a}\right)$$ + C
(d) None of these
(c) is correct Where c = Integration Constant

Question 11.
The value of $$\int_0^1 \frac{d x}{(1+x)(2+x)}$$ is : [1 Mark, Nov. 2007]
(a) log $$\frac{3}{4}$$
(b) log $$\frac{4}{3}$$
(c) log 12
(d) None of these
(b) is correct
$$\int_0^1 \frac{d x}{(1+x)(2+x)}$$
By Hit & Trial Method; we get
= $$\int_0^1\left(\frac{1}{1+x}-\frac{1}{2+x}\right)$$dx
= [log(1 + x)]10 – [log(2 + x)]10
= [log(1 + 1)- log(1 + 0)] – [log(2 + 1) – log(2 + 0)]
= log2 – 0 – log3 + log2
= 2log2 – log3
= log22 – log3 = log$$\frac{4}{3}$$ Question 12.
The value of $$\int_2^3$$(5 – x)dx – $$\int_2^3$$f(x)dx is:
(b) is correct. Question 13.
∫$$\frac{e^{\log _{e x} x} d x}{x}$$ is:
(a) x-1 + C
(b) x + c
(c) x2 + C
(d) None
(b) is correct
∫$$\frac{e^{\log _e x} d x}{x}$$ = ∫$$\frac{x}{x}$$
[Formula ax logab = bx]
= ∫dx = x + c

Question 14.
Evaluate ∫$$\frac{1}{(x-1)(x-2)}$$dx: [1 Mark, June 2008]
(a) log $$\left(\frac{x-2}{x-1}\right)$$ + C
(b) log[(x – 2)(x -1)] + C
(c) $$\left(\frac{x-1}{x-2}\right)$$ + C
(d) None
(a) is correct,
∫$$\frac{1}{(x-1)(x-2)}$$ = ∫$$\left(\frac{1}{(x-2)}-\frac{1}{x-1}\right)$$
(By Hit & Trial method)
= log(x – 2)- log(x -1)+ c
= log$$\left(\frac{x-2}{x-1}\right)$$ + c

Question 15.
$$\int_1^4$$(2x + 5) dx and the value is: [1 Mark, June 2008]
(a) 10
(b) 3
(c) 30
(d) None  ∫$$\frac{1}{x\left(x^5-1\right)}$$dx  