Integral Calculus Mcqs
Question 1.
∫10(ex + e-x) dx is:
(a) e – e-1
(b) e-1 – e
(c) e + e1
(d) None
Solution :
(a) is correctQuestion 2.
∫8x2(x3+2)3dx is equal to: [1 Mark, Nov. 2006]
(a) –43(x3 + 2) + C
(b) –43(x3 + 2)-2 + C
(c) 43(x3 + 2)2 + C
(d) None of these
Answer B] Is Correct
Question 3.
∫dxx2+a2√:
(a) 12log(x + x2+a2−−−−−−√) + C
(b) log(x + x2+a2−−−−−−√) + C
(c) log(xx2+a2−−−−−−√) + C
(d) 12log(xx2+a2−−−−−−√) + 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.
Question 4.
The value of ∫20x√x√+2−x√dx is: [1 Mark, Feb. 2007 & May 2007]
(a) 0
(b) 3
(c) 2
(d) 1
Answer:
(d) is correct
Question 5.
The Integral of (e3x + e-3x)/ex is: [1 Mark, May 2007]
(a) e2x2+e−4x4 + C
(b) e2x2−e−4x4 + C
(c) e2x – e-4x
(d) None of these
Answer:
(b) is correct
Where c = Integration (Arbitrary) Constant.
Question 6.
∫x2e3xdx is:
(a) x2.e3x – 2xe3x + 2e3x + C
(b) e3x3−x⋅e3x9 + 2e3x + C
(c) x2⋅e3x3−2x⋅e3x9+227e3x + C
(d) None of these
Answer:
(c) is correct.
Question 7.
∫212x1+x2dx: [1 Mark, May & Aug. 2007]
(a) loge52
(b) loge5 – loge2 + 1
(a) loge25
(d) None of these
Answer:
(a) is correct
Question 8.
The value of ∫e1(1+logx)xdx is: [Given Loge = 1]. [1 Mark, Aug. 2007]
(a) 1/2
(b) 3/2
(c) 1
(d) 5/2
Answer:
(b) is correct
Question 9.
Find ∫x3(x2+1)3dx: [1 Mark, Aug. 2007]
Answer:
(b) is correct
Question 10.
∫1x2−a2dx is: [1 Mark, Nov. 2007]
(a) log(x – a) – log(x + a) + C
(b) log x – ax+a + C
(c) 12a log(x−ax+a) + C
(d) None of these
Answer:
(c) is correct
Question 11.
The value of ∫10dx(1+x)(2+x) is :
(a) log 34
(b) log 43
(c) log 12
(d) None of these
Answer:
(b) is correct
∫10dx(1+x)(2+x)
By Hit & Trial Method; we get
= ∫10(11+x−12+x)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 = log43
Question 12.
The value of ∫32(5 – x)dx – ∫32f(x)dx is:
Answer:
(b) is correct.
Question 13.
∫elogexxdxx is:
(a) x-1 + C
(b) x + c
(c) x2 + C
(d) None
Answer:
(b) is correct
∫elogexdxx = ∫xx
[Formula ax logab = bx]
= ∫dx = x + c
Question 14.
Evaluate ∫1(x−1)(x−2)dx: [1 Mark, June 2008]
(a) log (x−2x−1) + C
(b) log[(x – 2)(x -1)] + C
(c) (x−1x−2) + C
(d) None
Answer:
(a) is correct,
∫1(x−1)(x−2) = ∫(1(x−2)−1x−1)
(By Hit & Trial method)
= log(x – 2)- log(x -1)+ c
= log(x−2x−1) + c
Question 15.
∫41(2x + 5) dx and the value is: [1 Mark, June 2008]
(a) 10
(b) 3
(c) 30
(d) None
Answer:
(c) Is correct.
= 15 + 15 = 30
Question 16.
∫1x(x5−1)dx
Answer:
(b) is correct