# Class 12 Maths MCQ – Fundamental Theorem of Calculus-1

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

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

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

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) $$\left (\frac{π}{2}-1\right)$$

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) $$\frac{e^4-1}{2}$$

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)

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

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

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

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

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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