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All questions of Integration- Definite Integrals for A Level Exam
Evaluate: 
- a)1/2
- b)1/4
- c)1
- d)1/8
Correct answer is option 'B'. Can you explain this answer?
Evaluate:
a)
1/2
b)
1/4
c)
1
d)
1/8
|
Sushil Kumar answered |
Let I = ∫(0 to pi/2)x3 sin(tan−1x4)/(1+x8)dx
Put tan−1 x4 = t
4x3/(1+x8)dx = dt
⇒x3dx/(1+x8) = dt/4
Now, I = 1/4∫sin t dt
=−[1/4 cos t]0 to pi/2 + c
= -1/4[cos(0) - cos(pi/2)] +c
= 1/4[-cos(pi/2) + cos(0)] +c
= ¼[-0+1]
= 1/4
Put tan−1 x4 = t
4x3/(1+x8)dx = dt
⇒x3dx/(1+x8) = dt/4
Now, I = 1/4∫sin t dt
=−[1/4 cos t]0 to pi/2 + c
= -1/4[cos(0) - cos(pi/2)] +c
= 1/4[-cos(pi/2) + cos(0)] +c
= ¼[-0+1]
= 1/4
Evaluate as limit of sum 
- a)20/5
- b)15/2
- c)20/3
- d)3/20
Correct answer is option 'C'. Can you explain this answer?
Evaluate as limit of sum
a)
20/5
b)
15/2
c)
20/3
d)
3/20
|
Om Desai answered |
∫(0 to 2)(x2 + x + 1)dx
= (0 to 2) [x3/3 + x2/2 + x]½
= [8/3 + 4/2 + 2]
= 40/6
= 20/3
= (0 to 2) [x3/3 + x2/2 + x]½
= [8/3 + 4/2 + 2]
= 40/6
= 20/3
The value of the integral 
is:- a)2e – 1
- b)2e + 1
- c)2e
- d)2(e – 1)
Correct answer is option 'D'. Can you explain this answer?
The value of the integral is:
is:a)
2e – 1
b)
2e + 1
c)
2e
d)
2(e – 1)
|
Infinity Academy answered |
Correct Answer : d
Explanation : ∫(-1 to 1) e|x| dx
∫(-1 to 0) e|x|dx + ∫(0 to 1) e|x|dx
e1 -1 + e1 - 1
=> 2(e - 1)
- a)-1
- b)zero
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
a)
-1
b)
zero
c)
1
d)
2
|
Praveen Kumar answered |
∫(0 to 4)(x)1/2 - x2 dx
= [[(x)3/2]/(3/2) - x2](0 to 4)
= [[2x3/2]/3 - x2](0 to 4)
= [[2(0)3/2]/3 - (0)2]] - [[2(4)3/2]/3 - (4)2]]
= 0-0
= 0
= [[(x)3/2]/(3/2) - x2](0 to 4)
= [[2x3/2]/3 - x2](0 to 4)
= [[2(0)3/2]/3 - (0)2]] - [[2(4)3/2]/3 - (4)2]]
= 0-0
= 0
If 
is- a)2/3
- b)4/5
- c)1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If is
isa)
2/3
b)
4/5
c)
1
d)
None of these
|
|
Om Desai answered |
In the question, it should be f’(2) instead of f”(2)
Explanation:- f(x) = ∫(0 to x) log(1+x2)
f’(x) = 2xdx/(1+x2)
f’(2) = 2(2)/(1+(2)2)
= 4/5
Explanation:- f(x) = ∫(0 to x) log(1+x2)
f’(x) = 2xdx/(1+x2)
f’(2) = 2(2)/(1+(2)2)
= 4/5
The value of 
is:- a)π/2
- b)1
- c)π/4
- d)π
Correct answer is option 'C'. Can you explain this answer?
The value of is:
is:a)
π/2
b)
1
c)
π/4
d)
π
|
|
Praveen Kumar answered |
1 + tan2 x = sec2 x
∫(π/2 to π/4)(sec2 x)/(1 + tan2 x)
=> ∫(π/2 to π/4)(sec2 x)/(sec2 x) dx
= ∫1(π/2 to π/4) dx
= [x](π/2 to π/4)
= π/2 - π/4 = π/4
∫(π/2 to π/4)(sec2 x)/(1 + tan2 x)
=> ∫(π/2 to π/4)(sec2 x)/(sec2 x) dx
= ∫1(π/2 to π/4) dx
= [x](π/2 to π/4)
= π/2 - π/4 = π/4
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
|
Sushil Kumar answered |
Option d is correct, because it is the property of definite integral
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx
Find 
- a)π/4
- b)π/2
- c)4π
- d)2π
Correct answer is option 'B'. Can you explain this answer?
Find
a)
π/4
b)
π/2
c)
4π
d)
2π
|
Vivek Patel answered |
Using trigonometric identities, we have
cos2x=cos2x-sin2x -(1) and cos2x+sin2x =1 -(2)
cos2x=1-sin2x , substituting this in equation (1) we get
cos2x=1-sin2x-sin2x=1-2sin2x
So,cos2x=1-2sin2x
2sin2x=1-cos2x
cos2x=cos2x-sin2x -(1) and cos2x+sin2x =1 -(2)
cos2x=1-sin2x , substituting this in equation (1) we get
cos2x=1-sin2x-sin2x=1-2sin2x
So,cos2x=1-2sin2x
2sin2x=1-cos2x
- a)π
- b)π/2
- c)2π
- d)π/4
Correct answer is option 'B'. Can you explain this answer?
a)
π
b)
π/2
c)
2π
d)
π/4
|
Tarun Kaushik answered |
For sin2(X), we will use the cos double angle formula:
cos(2X) = 1 - 2sin2(X)
The above formula can be rearranged to make sin2(X) the subject:
sin2(X) = 1/2(1 - cos(2X))
You can now rewrite the integration:
∫sin2(X)dX = ∫1/2(1 - cos(2X))dX
Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating:
1/2 x ∫(1 - cos(2X)) dX
= 1/2 x (X - 1/2sin(2X)) + C]-pi/4 to pi/4
∫sin2(X) dX = [1/2X - 1/4sin(2X)]-pi/4 to pi/4 + C
½[-pi/2] - 1/4sin(2(-pi/4)] - ½[pi/2] - 1/4sin(2(pi/4)]
= π/2
cos(2X) = 1 - 2sin2(X)
The above formula can be rearranged to make sin2(X) the subject:
sin2(X) = 1/2(1 - cos(2X))
You can now rewrite the integration:
∫sin2(X)dX = ∫1/2(1 - cos(2X))dX
Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating:
1/2 x ∫(1 - cos(2X)) dX
= 1/2 x (X - 1/2sin(2X)) + C]-pi/4 to pi/4
∫sin2(X) dX = [1/2X - 1/4sin(2X)]-pi/4 to pi/4 + C
½[-pi/2] - 1/4sin(2(-pi/4)] - ½[pi/2] - 1/4sin(2(pi/4)]
= π/2
- a)if f(2a – x) = – f(x)
- b)if f(2a – x) = f(x)
- c)if f(- x) = f(x)
- d)if f(- x) = – f(x)
Correct answer is option 'D'. Can you explain this answer?
a)
if f(2a – x) = – f(x)
b)
if f(2a – x) = f(x)
c)
if f(- x) = f(x)
d)
if f(- x) = – f(x)
|
Knowledge Hub answered |
∫(-a to a)f(x)dx
= ∫(0 to a) [f(x) + f(-x)] if f(x) is an odd function
⇒ f(-x) = -f(x)
= ∫(0 to a) [f(x) + f(-x)] if f(x) is an odd function
⇒ f(-x) = -f(x)
is: 
is:
The value of 
is:- a)10
- b)17/2
- c)7/2
- d)5
Correct answer is option 'A'. Can you explain this answer?
The value of is:
is:a)
10
b)
17/2
c)
7/2
d)
5
|
|
Sushil Kumar answered |
∫(-3 to 3) (x+1)dx
= ∫(-3 to -1) (x+1)dx + ∫(-1 to 3) (x+1) dx
= [x2 + x](-3 to -1) + [x2 + x](-1 to 3)
= [½ - 1 - (9/2 - 3)] + [9/2 + 3 - (½ - 1)]
= -[-4 + 2] + [4 + 4]
= -[-2] + [8]
= 10
If 
then the value of k is:- a)7/8
- b)5/8
- c)1/2
- d)3/2
Correct answer is option 'C'. Can you explain this answer?
If then the value of k is:
then the value of k is:a)
7/8
b)
5/8
c)
1/2
d)
3/2
|
|
Sushil Kumar answered |
Let I=∫(0 to k) 1/[1 + 4x2]dx = π8
Now, ∫(0 to k) 1/[4(1/4 + x2)]dx
= 2/4[tan−1 2x]0 to k
= 1/2tan-1 2k − 0 = π/8
1/2tan−1 2k = π8
⇒ tan−1 2k = π/4
⇒ 2k = 1
∴ k = 1/2
Now, ∫(0 to k) 1/[4(1/4 + x2)]dx
= 2/4[tan−1 2x]0 to k
= 1/2tan-1 2k − 0 = π/8
1/2tan−1 2k = π8
⇒ tan−1 2k = π/4
⇒ 2k = 1
∴ k = 1/2
The value of 
is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The value of is:
is:a)
b)
c)
d)
|
Samyak Jain answered |
1/2(cos π/2 + 1 -[cos(π/2-π/4)+1])=1/2+π/4
- a)g (x) h (s) = constant
- b)g (x) = h (x).
- c)g (x) - h (x) = constant
- d)h (x) + g (x) = constant
Correct answer is option 'C'. Can you explain this answer?
a)
g (x) h (s) = constant
b)
g (x) = h (x).
c)
g (x) - h (x) = constant
d)
h (x) + g (x) = constant
|
Swati Verma answered |
Since g(x) and h(x) are integrals of the same function , therefore ; g(x) – h(x) is constant.
- a)log 2
- b)0
- c)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
log 2
b)
0
c)
d)
none of these
|
Shilpa Saha answered |
∫ log(1-x) dx - ∫ log x dx = 0
Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.- a)10.19 km
- b)19.13 km
- c)15.13 km
- d)13.13 km
Correct answer is option 'D'. Can you explain this answer?
Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.
a)
10.19 km
b)
19.13 km
c)
15.13 km
d)
13.13 km
|
Rohit Yadav answered |
Add constant automatically
We know that,
We know that,
- a)– sin (log x)
- b)sin(log x)
- c)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
– sin (log x)
b)
sin(log x)
c)
d)
none of these
|
Learners Habitat answered |
Let t = log x (natural log). Then 
Hence, Option (b) is the correct answer.
- a)24/3
- b)25/3
- c)None of the above.
- d)26/3
Correct answer is option 'D'. Can you explain this answer?
a)
24/3
b)
25/3
c)
None of the above.
d)
26/3
|
Ajay Yadav answered |
The correct option is d : 26/3
- a)– log (1+ex)
- b)log (1+e−x)
- c)log(e−x+1)
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
a)
– log (1+ex)
b)
log (1+e−x)
c)
log(e−x+1)
d)
none of these
|
|
Divya Menon answered |
ANSWER :- a
Solution :- Let e^x/2 = tanθ. Then 1/2e^(x/2)dx = sec^2θdθ.
∫dx/(e^x+1) = 2∫½ e^(x/2)dx/[e^(x/2)(e^x+1)]
= 2∫[sec^2θdθ]/tanθ(sec^2θ)
= 2∫cosθ/sinθ)dθ = 2ln |sinθ|
From tanθ = e^(x/2) draw a right triangle to see that sinθ = e^x/2√e^x+1:
= 2ln∣e^pi/2√(e^x + 1)|
= ln |e^x/(e^x+1)|
= x−ln(e^x+1)+ C
is:
is equal to 
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