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) –$$\frac{124}{3}$$

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($$\sqrt{2}-\frac{1}{\sqrt{2}}$$)

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) $$\frac{11}{9} e^3-\frac{8}{9}$$

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-$$\frac{1}{3}$$

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) $$\frac{π}{6}$$

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

6. Find $$\int_3^45x^3 \,dx$$.
a) –$$\frac{185}{4}$$
b) –$$\frac{185}{3}$$
c) $$\frac{185}{2}$$
d) $$\frac{185}{4}$$

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) $$\frac{π}{6}$$

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) $$\frac{8\sqrt{2}+31}{4}$$

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) –$$\frac{5}{6}$$

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

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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