250+ TOP MCQs on Evaluation of Definite Integrals by Substitution | Class 12 Maths

Mathematics Online Quiz for IIT JEE Exam on “Evaluation of Definite Integrals by Substitution”.

1. Evaluate the integral (int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx).
a) 9
b) -9
c) (frac{9}{2})
d) –(frac{9}{2})
Answer: a
Clarification: I=(int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx)
Let (sqrt{x})=t
Differentiating both sides w.r.t x, we get
(frac{1}{2sqrt{x}} dx=dt)
The new limits are
When x=0 , t=0
When x=(frac{π^2}{4}, t=frac{π}{2})
∴(int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx=9int_0^{π/2} sin⁡t ,dt)
=(9[-cos⁡t]_0^{π/2})=-9(cos⁡ π/2-cos⁡0)=-9(0-1)=9

2. Find (int_0^1 20x^3 e^{x^4}) dx.
a) (e-1)
b) 5(e+1)
c) 5e
d) 5(e-1)
Answer: d
Clarification: I=(int_0^1 20x^3 e^{x^4}) dx
Let x⁴=t
Differentiating w.r.t x, we get
4x³ dx=dt
∴The new limits
When x=0, t=0
When x=1,t=1
∴(int_0^1 ,20x^3 ,e^{x^4} ,dx=int_0^1 5e^t dt)
(=5[e^t]_0^1=5(e^1-e^0))=5(e-1).

3. Find (int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx).
a) 4-(sqrt{2})
b) 4+2(sqrt{2})
c) 4-2(sqrt{2})
d) 1-2(sqrt{2})
Answer: c
Clarification: I=(int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx)
Let x⁵+3=t
Differentiating w.r.t x, we get
5x⁴ dx=dt
The new limits
when x=-1,t=2
when x=1,t=4
∴(int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx=int_2^4 frac{dt}{sqrt{t}})
=([2sqrt{t}]_2^4=2(sqrt{4}-sqrt{2})=4-2sqrt{2})

4. Find (int_0^{frac{sqrt{π}}{2}} 2x ,cos⁡ x^2 ,dx).
a) 1
b) (frac{1}{sqrt{2}})
c) –(frac{1}{sqrt{2}})
d) (sqrt{2})
Answer: b
Clarification: I=(int_0^{frac{sqrt{π}}{2}} ,2x ,cos⁡ x^2 ,dx)
Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
The new limits
When x=0,t=0
When (x={frac{sqrt{π}}{2}}, t=frac{π}{4})
∴(int_0^{frac{sqrt{π}}{2}} ,2x ,cos⁡ x^2 ,dx=int_0^{frac{π}{4}} ,cos⁡t ,dt)
(I =[sin⁡t]_0^{frac{π}{4}}=sin⁡ frac{π}{4}-sin⁡0=1/sqrt{2}).

5. Evaluate the integral (int_1^6 frac{sqrt{x}+3}{sqrt{x}} ,dx).
a) 9
b) (frac{9}{2})
c) –(frac{9}{2})
d) (frac{4}{5})
Answer: b
Clarification: I=(int_1^4 frac{sqrt{x}+3}{sqrt{x}} ,dx)
Let (sqrt{x}+3=t)
Differentiating w.r.t x, we get
(frac{1}{2sqrt{x}} ,dx=dt)
(frac{1}{sqrt{x}} ,dx=2 ,dt)
The new limits
When x=1,t=4
When x=4,t=5
∴(int_1^4 frac{sqrt{x}+3}{sqrt{x}} dx=int_4^5 ,t ,dt)
=([frac{t^2}{2}]_4^5=frac{5^2-4^2}{2}=frac{9}{2})

6. Find (int_1^2 frac{12 ,log⁡x}{x} ,dx).
a) -12 log⁡2
b) 24 log⁡2
c) 12 log⁡2
d) 24 log⁡4
Answer: b
Clarification: I=(int_1^2 frac{12 log⁡x}{x} ,dx)
Let log⁡x=t
Differentiating w.r.t x, we get
(frac{1}{x} ,dx=dt)
The new limits
When x=1,t=0
When x=2,t=log⁡2
(int_1^2 frac{12 log⁡x}{x} dx=12int_0^{log⁡2} ,t ,dt)
=(12[t^2]_0^{log⁡2}=12((log⁡2)^2-0))
=12 log⁡4=24 log⁡2(∵(log⁡2)²=log⁡2.log⁡2=log⁡4=2 log⁡2)

7. Find (int_0^{π/4} frac{5 ,sin⁡(tan^{-1}⁡x)}{1+x^2} ,dx).
a) 5-(frac{1}{sqrt{2}})
b) 5+(frac{5}{sqrt{2}})
c) -5+(frac{5}{sqrt{2}})
d) 5-(frac{5}{sqrt{2}})
Answer: d
Clarification: I=(int_0^1 frac{5 ,sin⁡(tan^{-1)}x}{1+x^2} ,dx)
Let tan^-1⁡x=t
Differentiating w.r.t x, we get
(frac{1}{1+x^2} ,dx=dt)
The new limits
When x=0, t=tan^-1⁡0=0
When x=1, t=tan^-1)1=π/4
∴(int_0^1 frac{5 ,sin⁡(tan^{-1}⁡x)}{1+x^2} ,dx=int_0^{π/4} ,5 ,sin⁡t ,dt)
=(5[-cos⁡t]_0^{π/4}=-5[cos⁡t]_0^{π/4})
(=-5(cos⁡ frac{π}{4}-cos⁡0)=-5(frac{1}{sqrt{2}-1})=5-frac{5}{sqrt{2}})

8. Find (int_{-1}^1 ,7x^6 ,(x^7+8)dx)
a) -386
b) –(frac{386}{3})
c) (frac{386}{3})
d) 386
Answer: c
Clarification: I=(int_{-1}^1 ,7x^6 ,(x^7+8)dx)
Let x⁷+8=t
Differentiating w.r.t x, we get
7x⁶ dx=dt
The new limits
When x=-1,t=7
When x=1,t=9
∴(int_{-1}^1 ,7x^6 ,(x^7+8)dx=int_7^9 ,t^2 ,dt)
=([frac{t^3}{3}]_7^9=frac{1}{3} (9^3-7^3)=frac{386}{3}).

9. Evaluate (int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx)
a) 14(3 log⁡2-1)
b) 14(3 log⁡2+1)
c) log⁡2-1
d) 3 log⁡2-1
Answer: a
Clarification: I=(int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx)
Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
The new limits
When x=(sqrt{2}), t=2
When x=2, t=4
∴(int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx =int_2^4 ,7 ,log⁡ t ,dt)
Using integration by parts, we get
(int_2^4 ,7 ,log⁡ t ,dt=7(log⁡ tint dt-int (log⁡t)’ int ,dt))
=7 (t log⁡t-t)₂⁴
=7(4 log⁡4-4-2 log⁡2+2)
=7(6 log⁡2-2)=14(3 log⁡2-1)

10. Find (int_2^3 ,2x^2 ,e^{x^3} ,dx).
a) (e^{27}-e^8)
b) (frac{2}{3} (e^{27}-e^8))
c) (frac{2}{3} (e^8-e^{27}))
d) (frac{2}{3} (e^{27}+e^8))
Answer: b
Clarification: I=(int_2^3 ,2x^2 ,e^{x^3} ,dx)
Let x³=t
Differentiating w.r.t x, we get
3x² dx=dt
x² dx=(frac{dt}{3})
The new limits
When x=2, t=8
When x=3, t=27
∴(int_2^3 ,2x^2 ,e^{x^3} ,dx=frac{2}{3} int_8^{27} ,e^t ,dt)
=(frac{2}{3} [e^t]_8^{27}=frac{2}{3} (e^{27}-e^8).)

Mathematics Online Quiz for IIT JEE Exam,