This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) 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) –\(\frac{9}{2}\)
View Answer
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=\(\frac{π^2}{4}, t=\frac{π}{2}\)
∴\(\int_0^{\frac{π^2}{4}} \frac{9 sin\sqrt{x}}{2\sqrt{x}} dx=9\int_0^{π/2} sint \,dt\)
=\(9[-cost]_0^{π/2}\)=-9(cos π/2-cos0)=-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)
View Answer
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\(\sqrt{2}\)
View Answer
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
∴\(\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx=\int_2^4 \frac{dt}{\sqrt{t}}\)
=\([2\sqrt{t}]_2^4=2(\sqrt{4}-\sqrt{2})=4-2\sqrt{2}\)
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) \(\sqrt{2}\)
View Answer
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 \(x={\frac{\sqrt{π}}{2}}, t=\frac{π}{4}\)
∴\(\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos x^2 \,dx=\int_0^{\frac{π}{4}} \,cost \,dt\)
\(I =[sint]_0^{\frac{π}{4}}=sin \frac{π}{4}-sin0=1/\sqrt{2}\).
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) \(\frac{4}{5}\)
View Answer
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 \(\int_1^2 \frac{12 \,logx}{x} \,dx\).
a) -12 log2
b) 24 log2
c) 12 log2
d) 24 log4
View Answer
Explanation: I=\(\int_1^2 \frac{12 logx}{x} \,dx\)
Let logx=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=log2
\(\int_1^2 \frac{12 logx}{x} dx=12\int_0^{log2} \,t \,dt\)
=\(12[t^2]_0^{log2}=12((log2)^2-0)\)
=12 log4=24 log2(∵(log2)2=log2.log2=log4=2 log2)
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-\(\frac{5}{\sqrt{2}}\)
View Answer
Explanation: I=\(\int_0^1 \frac{5 \,sin(tan^{-1)}x}{1+x^2} \,dx\)
Let tan-1x=t
Differentiating w.r.t x, we get
\(\frac{1}{1+x^2} \,dx=dt\)
The new limits
When x=0, t=tan-10=0
When x=1, t=tan-1)1=π/4
∴\(\int_0^1 \frac{5 \,sin(tan^{-1}x)}{1+x^2} \,dx=\int_0^{π/4} \,5 \,sint \,dt\)
=\(5[-cost]_0^{π/4}=-5[cost]_0^{π/4}\)
\(=-5(cos \frac{π}{4}-cos0)=-5(\frac{1}{\sqrt{2}-1})=5-\frac{5}{\sqrt{2}}\)
8. Find \(\int_{-1}^1 \,7x^6 \,(x^7+8)dx\)
a) -386
b) –\(\frac{386}{3}\)
c) \(\frac{386}{3}\)
d) 386
View Answer
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 log2-1)
b) 14(3 log2+1)
c) log2-1
d) 3 log2-1
View Answer
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
\(\int_2^4 \,7 \,log t \,dt=7(log t\int dt-\int (logt)’ \int \,dt)\)
=7 (t logt-t)24
=7(4 log4-4-2 log2+2)
=7(6 log2-2)=14(3 log2-1)
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) \(\frac{2}{3} (e^{27}+e^8)\)
View Answer
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
∴\(\int_2^3 \,2x^2 \,e^{x^3} \,dx=\frac{2}{3} \int_8^{27} \,e^t \,dt\)
=\(\frac{2}{3} [e^t]_8^{27}=\frac{2}{3} (e^{27}-e^8).\)
Sanfoundry Global Education & Learning Series – Mathematics – Class 12.
To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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