Mathematics Questions and Answers – Fundamental Theorem of Calculus-2

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This set of Mathematics Multiple Choice Questions and Answers for Class 12 focuses on “Fundamental Theorem of Calculus-2”.

1. Evaluate the integral \(\int_1^5x^2 \,dx\).
a) \(\frac{125}{3}\)
b) \(\frac{124}{3}\)
c) 124
d) –\(\frac{124}{3}\)
View Answer

Answer: b
Explanation: Let I=\(\int_1^5x^2 \,dx\)
F(x)=\(\int x^2 \,dx\)
=\(\frac{x^3}{3}\)
By using the fundamental theorem of calculus, we get
I=F(5)-F(1)
=\(\frac{5^3}{3}-\frac{1^3}{3}=\frac{125-1}{3}=\frac{124}{3}\)
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2. Find \(\int_{π/4}^{π/2}7 \,cos⁡x \,dx\).
a) 7(1-\(\frac{1}{\sqrt{2}}\))
b) -7(1-\(\frac{1}{\sqrt{2}}\))
c) 7(1+\(\frac{1}{\sqrt{2}}\))
d) 7(\(\sqrt{2}-\frac{1}{\sqrt{2}}\))
View Answer

Answer: a
Explanation: Let \(I=\int_{π/4}^{π/2}7 \,cos⁡x \,dx\)
F(x)=∫ 7 cos⁡x dx
=7(sin⁡x)
Applying the limits by using the second fundamental theorem of calculus, we get
\(I=F(\frac{π}{2})-F(\frac{π}{4})=7(sin\frac{π}{2}-sin⁡ \frac{π}{4})=7(1-\frac{1}{\sqrt{2}})\)

3. The value of the integral \(\int_0^1(x+3) \,e^{3x} \,dx\).
a) \(\frac{8e^3}{9}\)
b) \(\frac{11}{9} e^3-8\)
c) \(\frac{e^{3x}}{9}(x+8)\)
d) \(\frac{11}{9} e^3-\frac{8}{9}\)
View Answer

Answer: d
Explanation: Let \(I=\int_0^1(x+3) \,e^{3x} \,dx\)
F(x)=\(\int (x+3) \,e^{3x} \,dx\)
By using the formula \(\int \,u.v \,dx=u\int \,v dx-\int \,u'(\int \,v \,dx)\), we get
F(x)=(x+3) \(\int e^{3x} \,dx-\int \,(x+3)’\int \,e^{3x} \,dx\)
=\(\frac{(x+3) \,e^{3x}}{3}-\int \frac{e^{3x}}{3} dx\)
=\(\frac{(x+3) e^{3x}}{3}-\frac{e^{3x}}{9}\)
=\(\frac{e^{3x}}{3} (x+3-\frac{1}{3})=\frac{e^{3x}}{9}(3x+8)\)
Applying the limits, we get
I=F(1)-F(0)
=\(\frac{e^{3(1)}}{9} (3+8)-\frac{e^{3(0)}}{9}(0+8)\)
=\(\frac{11}{9} e^3-\frac{8}{9}\).
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4. Find \(\int_0^π(1-sin⁡3x)dx\).
a) \(\frac{3π-2}{4}\)
b) 3π-1
c) \(\frac{3π-2}{3}\)
d) π-\(\frac{1}{3}\)
View Answer

Answer: c
Explanation: Let \(I=\int_0^π(1-sin⁡3x)dx\)
F(x)=∫ 1-sin⁡3x dx
=x+\(\frac{cos⁡3x}{3}\)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(π)-F(0)
=\(π+\frac{cos⁡3π}{3}-0-\frac{cos⁡0}{3}\)
=\(π-\frac{1}{3}-\frac{1}{3}=π-\frac{2}{3}=\frac{3π-2}{3}\).

5. Evaluate the integral \(\int_1^{\sqrt{3}} \frac{3}{1+x^2}\).
a) \(\frac{π}{2}\)
b) \(\frac{π}{4}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{6}\)
View Answer

Answer: b
Explanation: Let \(I=\int_1^{√3} \frac{3}{1+x^2}\)
F(x)=\(\int \frac{3}{1+x^2}dx\)
=3\(\int \frac{1}{1+x^2} \,dx\)
=3 tan-1⁡x
Applying the limits, we get
I=F(\(\sqrt{3}\))-F(1)
=3 tan-1⁡\(\sqrt{3}\)-3 tan-1⁡1
\(3(\frac{π}{3})-\frac{3π}{4}=\frac{4π-3π}{4}=\frac{π}{4}\).
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6. Find \(\int_3^45x^3 \,dx\).
a) –\(\frac{185}{4}\)
b) –\(\frac{185}{3}\)
c) \(\frac{185}{2}\)
d) \(\frac{185}{4}\)
View Answer

Answer: d
Explanation: Let \(I=\int_3^45x^3 \,dx\)
F(x)=∫ 5x3 dx
=\(\frac{5x^4}{4}\)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(4)-F(3)
=\(\frac{5(4)^3}{4}-\frac{5(3)^3}{4}=\frac{5}{4}(4^3-3^3)\)
=\(\frac{5}{4} (64-27)=\frac{5}{4} (37)=\frac{185}{4}\)

7. Evaluate the definite integral \(\int_0^1 sin^2⁡x \,dx\).
a) –\(\frac{π}{2}\)
b) π
c) \(\frac{π}{4}\)
d) \(\frac{π}{6}\)
View Answer

Answer: c
Explanation: Let \(I=\int_0^{π/2}sin^{2⁡}x \,dx\)
F(x)=\(\int sin^2⁡x \,dx\)
=\(\int \frac{(1-cos⁡2x)}{2} \,dx\)
=\(\frac{1}{2} (x-\frac{sin⁡2x}{2})\)
Applying the limits, we get
\(I=F(\frac{π}{2})-F(0)=\frac{1}{2} (\frac{π}{2}-\frac{sin⁡π}{2})-\frac{1}{2} (0-\frac{sin⁡0}{2})\)
=\(\frac{1}{4} (π-0)-0=\frac{π}{4}\).
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8. Find \(\int_1^2\sqrt{x}-3x \,dx\).
a) \(\frac{8\sqrt{2}-31}{6}\)
b) \(8\sqrt{2}-31\)
c) \(\frac{\sqrt{2}-31}{3}\)
d) \(\frac{8\sqrt{2}+31}{4}\)
View Answer

Answer: a
Explanation: Let \(I=\int_1^2 \sqrt{x}-3x \,dx\)
F(x)=\(\int \sqrt{x}-3x \,dx\)
=\(\frac{x^{1/2+1}}{1/2+1}-\frac{3x^2}{2}=\frac{2x^{\frac{3}{2}}}{3}-\frac{3x^2}{2}\)
By using the second fundamental theorem of calculus, we get
I=F(2)-F(1)=\(\left(\frac{2×2^{3/2}}{3}-\frac{3×2^2}{2}\right)-\left(\frac{2×1^{3/2}}{3}-\frac{3×1^2}{2}\right)\)
I=\(\frac{4\sqrt{2}}{3}-6-\frac{2}{3}+\frac{3}{2}=\frac{8\sqrt{2}-36-4+9}{6}=\frac{8\sqrt{2}-31}{6}\)

9. Find the value \(\int_{-1}^23x+x^2-2 \,dx\).
a) –\(\frac{4}{3}\)
b) \(\frac{3}{2}\)
c) \(\frac{5}{6}\)
d) –\(\frac{5}{6}\)
View Answer

Answer: b
Explanation: Let \(I=\int_{-1}^23x+x^2-2 \,dx\)
F(x)=\(\int 3x+x^2-2 \,dx\)
=\(\frac{3x^2}{2}+\frac{x^3}{3}-2x\)
Applying the limits, we get
I=F(2)-F(-1)
I=\(\left(\frac{(3×2^3)}{2}+\frac{2^3}{3}-2(2)\right)-\left(\frac{3 (-1)^2}{2}+\frac{(-1)^3}{3}-2(-1)\right)\)
I=\(6+\frac{8}{3}-4-\frac{3}{2}+\frac{1}{3}-2=\frac{3}{2}\).
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10. Find \(\int_1^2 log⁡x.x^2 dx\)
a) log⁡2-\(\frac{7}{3}\)
b) \(\frac{8}{3}\) log⁡2-5
c) \(\frac{8}{3}\) log⁡2-log⁡3
d) \(\frac{8}{3}\) log⁡2
View Answer

Answer: b
Explanation: \(I=\int_0^1 log⁡x.x^2 dx\)
F(x)=\(\int log⁡x.x^2 dx\)
By using the formula \(\int u.v dx=u\int v dx-\int u'(\int v dx)\), we get
\(\int log⁡x.x^2 \,dx=log⁡x \int x^2 dx-\int (log⁡x)’ \int \,x^2 dx\)
=\(\frac{x^3 log⁡x}{3}-\int \frac{1}{x}.x^3/3 dx\)
∴\(F(x)=\frac{x^3 log⁡x}{3}-\frac{x^3}{9}=\frac{x^3}{3} (log⁡x-\frac{1}{3})\)
Hence, by using the fundamental theorem of calculus, we get
I=F(2)-F(1)
I=\(\frac{2^3}{3} \,(log⁡2-\frac{2}{3})-\frac{1^3}{3} \,(log⁡1-\frac{1}{3})\)
I=\(\frac{2^3}{3} \,log⁡2-\frac{16}{3}+\frac{1}{3}\)
I=\(\frac{8}{3}\) log⁡2-5

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter