# Mathematics Questions and Answers – Evaluation of Definite Integrals by Substitution

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

1. Evaluate the integral $$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx$$.
a) 9
b) -9
c) $$\frac{9}{2}$$
d) –$$\frac{9}{2}$$

Explanation: I=$$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx$$
Let $$\sqrt{x}$$=t
Differentiating both sides w.r.t x, we get
$$\frac{1}{2\sqrt{x}} dx=dt$$
The new limits are
When x=0 , t=0
When x=$$\frac{π^2}{4}, t=\frac{π}{2}$$
∴$$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx=9\int_0^{π/2} sin⁡t \,dt$$
=$$9[-cos⁡t]_0^{π/2}$$=-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)

Explanation: 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$$\sqrt{2}$$

Explanation: 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
∴$$\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx=\int_2^4 \frac{dt}{\sqrt{t}}$$
=$$[2\sqrt{t}]_2^4=2(\sqrt{4}-\sqrt{2})=4-2\sqrt{2}$$
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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) $$\sqrt{2}$$

Explanation: 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) $$\frac{4}{5}$$

Explanation: I=$$\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx$$
Let $$\sqrt{x}+3=t$$
Differentiating w.r.t x, we get
$$\frac{1}{2\sqrt{x}} \,dx=dt$$
$$\frac{1}{\sqrt{x}} \,dx=2 \,dt$$
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

Explanation: 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
$$\int_1^2 \frac{12 log⁡x}{x} dx=12\int_0^{log⁡2} \,t \,dt$$
=$$12[t^2]_0^{log⁡2}=12((log⁡2)^2-0)$$
=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-$$\frac{5}{\sqrt{2}}$$

Explanation: 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

Explanation: 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

Explanation: 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
$$\int_2^4 \,7 \,log⁡ t \,dt=7(log⁡ t\int dt-\int (log⁡t)’ \int \,dt)$$
=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) $$\frac{2}{3} (e^{27}+e^8)$$

Explanation: 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
∴$$\int_2^3 \,2x^2 \,e^{x^3} \,dx=\frac{2}{3} \int_8^{27} \,e^t \,dt$$
=$$\frac{2}{3} [e^t]_8^{27}=\frac{2}{3} (e^{27}-e^8).$$

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