Fundamental Theorem of Calculus - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 7 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)
∴\(\int_0^8x \,dx=\frac{8^2}{2}-0=32\)

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.

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.

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=\(\frac{8}{3} log⁡2-\frac{8}{3}-0+\frac{1}{3}=\frac{8}{3} log⁡2-\frac{7}{3}\).

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=\(\frac{4}{e}\).

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