Mathematics Questions and Answers – Fundamental Theorem of Calculus-1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Fundamental Theorem of Calculus-1”.

1. Find \(\int_0^8x \,dx\).
a) 32
b) 34
c) 21
d) 24
View Answer

Answer: a
Explanation: Let I=\(\int_0^8x \,dx\)
F(x)=\(\int x \,dx=\frac{x^2}{2}\)
Using the second fundamental theorem of calculus, we get
I=F(8)-f(0)
∴\(\int_0^8x \,dx=\frac{8^2}{2}-0=32\)
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2. Find \(\int_0^{\frac{π}{2}} \,5 \,sin⁡x \,dx\).
a) -5
b) 9
c) 5
d) -9
View Answer

Answer: c
Explanation: Let \(I=\int_0^{\frac{π}{2}} \,5 \,sin⁡x \,dx\)
F(x)=\(\int5 \,sin⁡x \,dx=-5 \,cos⁡x\)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(\(\frac{π}{2}\))-F(0)
∴\(\int_0^{\frac{π}{2}} \,5 \,sin⁡x \,dx=-5[cos⁡\frac{π}{2}-cos⁡0]\)
=-5[0-1]=5

3. Find the value of \(\int_4^5 \,log⁡x \,dx\).
a) 5 log⁡5-log⁡4+1
b) 5 log⁡5-4 log⁡4-1
c) 4 log⁡5-4 log⁡4-1
d) 5-4 log⁡4-log⁡5
View Answer

Answer: b
Explanation: Let I=\(\int_4^5 \,log⁡x \,dx\).
F(x)=∫ log⁡x dx
By using the formula \(\int \,u.v dx=u \int v \,dx-\int u'(\int \,v \,dx)\), we get
\(\int log ⁡x \,dx=log⁡x \int \,dx-\int(log⁡x)’\int \,dx\)
F(x)=x log⁡x-∫ dx=x(log⁡x-1).
Applying the limits using the fundamental theorem of calculus, we get
I=F(5)-F(4)=(5 log⁡5-5)-(4 log⁡4-4)
=5 log⁡5-4 log⁡4-1.
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4. Find \(\int_0^{\frac{π}{4}} \,9 \,cos^2⁡x \,dx\).
a) \(\frac{9}{2}\left (\frac{π}{6}-1\right)\)
b) \(\frac{9}{4}\left (\frac{π}{2}+1\right)\)
c) \(\frac{9}{4}\left (\frac{π}{2}-1\right)\)
d) \(\left (\frac{π}{2}-1\right)\)
View Answer

Answer: c
Explanation: Let I=\(\int_0^{\frac{π}{4}} \,9 \,cos^2⁡x \,dx\).
F(x)=\(\int \,9 \,cos^2⁡x \,dx\)
=9\(\int(\frac{1+cos⁡2x}{2})dx\)
=\(\frac{9}{2} (x-\frac{sin⁡2x}{2})\)
Applying the limits, we get
I=\(F(\frac{π}{4})-F(0)=\frac{9}{2} \left (\frac{π}{4}-\frac{sin⁡2(\frac{π}{4})}{2}\right)-\frac{9}{2} (0-\frac{sin⁡0}{2})\)
=\(\frac{9}{2}\left (\frac{π}{4}-\frac{sin⁡π/2}{2}\right )=\frac{9}{4} (π/2-1)\)

5. Find \(\int_0^2 \,e^{2x} \,dx\).
a) \(\frac{e^4-1}{6}\)
b) \(\frac{e^4+1}{2}\)
c) \(\frac{e-1}{2}\)
d) \(\frac{e^4-1}{2}\)
View Answer

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

Answer: a
Explanation: Let \(I=\int_{π/4}^{π/2} \,2sinx \,sin⁡(cos⁡x) \,dx\)
F(x)=\(\int 2 \,sin⁡x \,sin⁡(cos⁡x)dx\)
Let cos⁡x=t
Differentiating w.r.t x, we get
sin⁡x dx=dt
∴\(\int 2 \,sin⁡x \,sin⁡(cos⁡x)dx=\int 2 \,sin⁡t \,dt=-2 \,cos⁡t\)
Replacing t with cos⁡x, we get
∴∫ 2 sin⁡x sin⁡(cos⁡x)dx=-2 cos⁡(cos⁡x)
By applying the limits, we get
\(I=F(\frac{π}{4})-F(\frac{π}{2})=-2 cos⁡(\frac{cos⁡π}{4})+2 cos⁡(\frac{cos⁡π}{2})\)
=2(1-cos⁡\(\frac{1}{\sqrt{2}}\))

7. Find \(\int_{-2}^1 \,5x^4 \,dx\).
a) 54
b) 75
c) 33
d) 36
View Answer

Answer: c
Explanation: \(I=\int_{-2}^1 \,5x^4 \,dx\)
F(x)=\(\int5x^4 \,dx=5(\frac{x^5}{5})=x^5\)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(1)-F(-2)
=(1)5-(-2)5=1+32=33.
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8. Find \(\int_0^3 \,e^x \,dx\).
a) e3+1
b) -e3-1
c) e3-1
d) 3e3-2
View Answer

Answer: c
Explanation: Let I=\(\int_0^3 \,e^x \,dx\)
F(x)=\(\int \,e^x \,dx=e^x\)
Applying the limits, we get
I=F(3)-F(0)
=e3-e0=e3-1.

9. Find \(\int_0^{π/4} \,2 \,tan⁡x \,dx\).
a) log⁡2
b) log⁡\(\sqrt{2}\)
c) 2 log⁡2
d) 0
View Answer

Answer: a
Explanation: \(I=\int_0^{π/4} \,2 \,tan⁡x \,dx\)
F(x)=∫ 2 tan⁡x dx
=2∫ tan⁡x dx
=2 log⁡|sec⁡x|
Therefore, by using the fundamental theorem of calculus, we get
I=F(π/4)-F(0)
\(=2\left(log⁡|sec \frac{⁡π}{4}|-log⁡|sec⁡0|\right)=2 log⁡\sqrt{2}-log⁡1\)
\(=2 log⁡\sqrt{2}=log⁡(\sqrt{2})^2=log⁡2\)
I=\(\frac{8}{3} log⁡2-\frac{8}{3}-0+\frac{1}{3}=\frac{8}{3} log⁡2-\frac{7}{3}\).
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10. Find \(\int_{-1}^1 \,2xe^x \,dx\).
a) \(\frac{4}{e}\)
b) 4e
c) –\(\frac{4}{e}\)
d) -4e
View Answer

Answer: a
Explanation: \(I=\int_{-1}^1 \,2xe^x \,dx\)
F(x)=\(\int 2xe^x dx\)
By using the formula, \(\int u.v \,dx=u \int v \,dx-\int u'(\int v \,dx)\)
F(x)=2x\(\int e^x dx-\int(2x)’\int e^x \,dx\)
=\(2xe^x-\int 2e^x dx\)
=\(2e^x (x-1)\)
Therefore, by using the fundamental theorem of calculus, we get
I=F(1)-F(-1)
I=2e1 (1-1)-2e-1 (-1-1)
I=\(\frac{4}{e}\).

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