Fundamental Theorem of Calculus 2 - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) 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)
=\(\frac{5^3}{3}-\frac{1^3}{3}=\frac{125-1}{3}=\frac{124}{3}\)

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)
=\(\frac{e^{3(1)}}{9} (3+8)-\frac{e^{3(0)}}{9}(0+8)\)
=\(\frac{11}{9} e^3-\frac{8}{9}\).

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
\(3(\frac{π}{3})-\frac{3π}{4}=\frac{4π-3π}{4}=\frac{π}{4}\).

Subscribe Now: Class 12 - Mathematics Newsletter | Important Subjects Newsletters

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
\(I=F(\frac{π}{2})-F(0)=\frac{1}{2} (\frac{π}{2}-\frac{sin⁡π}{2})-\frac{1}{2} (0-\frac{sin⁡0}{2})\)
=\(\frac{1}{4} (π-0)-0=\frac{π}{4}\).

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=\(6+\frac{8}{3}-4-\frac{3}{2}+\frac{1}{3}-2=\frac{3}{2}\).

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.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

« Prev - Class 12 Maths MCQ – Fundamental Theorem of Calculus-1

» Next - Class 12 Maths MCQ – Evaluation of Definite Integrals by Substitution

Related Posts:

Recommended Articles: