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
Integral Calculus – CA Foundation Maths Study Material 1
= 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
Answer:
Integral Calculus – CA Foundation Maths Study Material 2

Integral Calculus – CA Foundation Maths Study Material

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
Answer:
(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.
Integral Calculus – CA Foundation Maths Study Material 3

III Detail Method
Let t – x = \(\sqrt{x^2+a^2}\)
Integral Calculus – CA Foundation Maths Study Material 4

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
Answer:
(d) is correct
Integral Calculus – CA Foundation Maths Study Material 5
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
Answer:
(b) is correct
Integral Calculus – CA Foundation Maths Study Material 6
Where c = Integration (Arbitrary) Constant.

Integral Calculus – CA Foundation Maths Study Material

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
Answer:
(c) is correct.
By parts; we get
Integral Calculus – CA Foundation Maths Study Material 7

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
Answer:
(a)
Integral Calculus – CA Foundation Maths Study Material 8
= 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
Answer:
(b) is correct
Integral Calculus – CA Foundation Maths Study Material 9

Question 9.
Find ∫\(\frac{x^3}{\left(x^2+1\right)^3}\)dx: [1 Mark, Aug. 2007]
Integral Calculus – CA Foundation Maths Study Material 10
Answer:
(b) is correct
Integral Calculus – CA Foundation Maths Study Material 11

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
Answer:
(c) is correct
Integral Calculus – CA Foundation Maths Study Material 12
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
Answer:
(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\(\frac{4}{3}\)

Integral Calculus – CA Foundation Maths Study Material

Question 12.
The value of \(\int_2^3\)(5 – x)dx – \(\int_2^3\)f(x)dx is:
Answer:
(b) is correct.
Integral Calculus – CA Foundation Maths Study Material 13

Question 13.
∫\(\frac{e^{\log _{e x} x} d x}{x}\) is:
(a) x-1 + C
(b) x + c
(c) x2 + C
(d) None
Answer:
(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
Answer:
(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
Answer:
(c) Is correct.
Integral Calculus – CA Foundation Maths Study Material 14
= 15 + 15 = 30

Integral Calculus – CA Foundation Maths Study Material

Question 16.
∫\(\frac{1}{x\left(x^5-1\right)}\)dx
Integral Calculus – CA Foundation Maths Study Material 15
Answer:
(b) is correct
Integral Calculus – CA Foundation Maths Study Material 16

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