Improper Integrals Questions and Answers

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Improper Integrals – 2”.

1. Find the value of ∫tan^-1⁡(x)dx.
a) sec^-1 (x) – ¹⁄₂ ln⁡(1 + x²)
b) xtan^-1 (x) – ¹⁄₂ ln⁡(1 + x²)
c) xsec^-1 (x) – ¹⁄₂ ln⁡(1 + x²)
d) tan^-1 (x) – ¹⁄₂ ln⁡(1 + x²)
View Answer

Answer: b
Explanation: Add constant automatically
Given, ∫tan^-1⁡(x)dx
Putting, x = tan(y),
We get, dy = sec²(y)dy,
∫ysec²(y)dy
By integration by parts,
ytan(y) – log⁡(sec⁡(y)) = xtan^-1 (x) – ¹⁄₂ ln⁡(1 + x²).

2. Integration of (Sin(x) + Cos(x))e^x is?
a) e^x Cos(x)
b) e^x Sin(x)
c) e^x Tan(x)
d) e^x (Sin(x) + Cos(x))
View Answer

Answer: b
Explanation: Add constant automatically
Let f(x) = e^x Sin(x)
∫e^x Sin(x)dx = e^x Sin(x) – ∫e^x Cos(x)dx
∫e^x Sin(x)dx + ∫e^x Cos(x)dx = ∫e^x [Cos(x) + Sin(x)]dx = e^x Sin(x).

3. Find the value of ∫x³ Sin(x)dx.
a) x³ Cos(x) + 3x² Sin(x) + 6xCos(x) – 6Sin(x)
b) – x³ Cos(x) + 3x² Sin(x) – 6Sin(x)
c) – x³ Cos(x) – 3x² Sin(x) + 6xCos(x) – 6Sin(x)
d) – x³ Cos(x) + 3x² Sin(x) + 6xCos(x) – 6Sin(x)
View Answer

Answer: d
Explanation: Add constant automatically
Let f(x) = x³ Sin(x)
∫x³ Sin(x)dx = – x³ Cos(x) + 3∫x² Cos(x)dx
∫x² Cos(x)dx = x² Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = – xCos(x) + ∫Cos(x)dx = – xCos(x) + Sin(x)
=> ∫x³ Sin(x)dx = – x³ Cos(x) + 3[x² Sin(x) – 2[ – xCos(x) + Sin(x)]]
=> ∫x³ Sin(x)dx = – x³ Cos(x) + 3x² Sin(x) + 6xCos(x) – 6Sin(x).

4. Value of ∫uv dx,where u and v are function of x.
a) \(\sum_{i=1}^n(-1)^i u_i v^{i+1}\)
b) \(\sum_{i=0}^nu_i v^{i+1}\)
c) \(\sum_{i=0}^n(-1)^i u_i v^{i+1}\)
d) \(\sum_{i=0}^n(-1)^i u_i v^{n-i}\)
View Answer

Answer: c
Explanation: Add constant automatically
Given, f(x)=\(\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}\)

5. Find the value of ∫x⁷ Cos(x) dx.
a) x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x)
b) x⁷ Sin(x) – 7x⁶ Cos(x) + 42x⁵ Sin(x) – 210x⁴ Cos(x) + 840x³ Sin(x) – 2520x² Cos(x) + 5040xSin(x) – 5040Cos(x)
c) x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x)
d) x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 10080Cos(x)
View Answer

Answer: a
Explanation: Add constant automatically
By, f(x)=\(\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}\)
Let, u = x⁷ and v = Cos(x),
∫x⁷ Cos(x) dx = x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x)

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6. Find the value of ∫x³ e^x e^2x e^3x….e^nx dx.
a) \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
b) \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
c)\(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
d)\(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
View Answer

Answer: a
Explanation: Add constant automatically
By, f(x)=\(\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}\)
Let, u = x³ and v=e^x e^2x e^3x…..e^nx=e^{x(1+2+3+…n)}=\(e^{\frac{n(n+1)x}{2}}\),
\(\int x^3 e^x e^2x e^3x……..e^nx dx\)
\(=x^3 \frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x}+3x^2 [\frac{2}{n(n+1)}]^2 e^{\frac{n(n+1)}{2}x}\)
\(+6x[\frac{2}{n(n+1)}]^3 e^{\frac{n(n+1)}{2}x}+6[\frac{2}{n(n+1)}]^4 e^{\frac{n(n+1)}{2}x}\)
=\(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2+6[\frac{2}{n(n+1)}]^3\right]\)

7. Find the area of a function f(x) = x² + xCos(x) from x = 0 to a, where, a>0.
a) ^a²⁄₂ + aSin(a) + Cos(a) – 1
b) ^a³⁄₃ + aSin(a) + Cos(a)
c) ^a³⁄₃ + aSin(a) + Cos(a) – 1
d) ^a³⁄₃ + Cos(a) + Sin(a) – 1
View Answer

Answer: c
Explanation: Given, f(x) = x² + xCos(x)
Hence, F(x) = ∫x² + xCos(x) dx = ^x³⁄₃ + xSin(x) + Cos(x)
Hence, area inside f(x) is,
F(a) – F(0) = ^a³⁄₃ + aSin(a) + Cos(a) – 1.

8. Find the area ^ln(x)⁄_x from x = x = ae^b to a.
a) ^b²⁄₂
b) ^b⁄₂
c) b
d) 1
View Answer

Answer: a
Explanation:
Let, F(x)=\(\int \frac{ln⁡(x)}{x} dx\)
Let, z=ln⁡(x)=>dz=dx/x
=F(x)=∫ zdz=\(\frac{z^2}{2}=\frac{ln^2⁡(x)}{2}\)
Area inside curve from 4a to a is,
\(F(ae^b)-F(a)=\frac{ln^2⁡(ae^b )}{2}-\frac{ln^2⁡(a)}{2}=\frac{ln^2⁡(\frac{ae^b}{a})}{2}=\frac{ln^2⁡(e^b)}{2}=\frac{b}{2}\)

9. Find the area inside a function f(t) = \( \frac{t}{(t+3)(t+2)} dt\) from t = -1 to 0.
a) 4 ln⁡(3) – 5ln⁡(2)
b) 3 ln⁡(3)
c)3 ln⁡(3) – 4ln⁡(2)
d) 3 ln⁡(3) – 5 ln⁡(2)
View Answer

Answer: d
Explanation:
Now, F(t)=\(\int \frac{t}{(t+3)(t+2)} dt\)
F(t)=\(\int \frac{t}{(t+3)(t+2)} dt\)
=\(\int [\frac{3}{t+3}-\frac{2}{t+2}]dx\)
=\(\int [\frac{3}{t+3}]dx-\int [\frac{2}{t+2}]dx\)
=3 ln⁡(t+3)-2ln⁡(t+2)
Now area inside a function is, F(0) – F(-1),
hence, F(0)-F(-1)=3 ln⁡(3)-2 ln⁡(2)-3 ln⁡(2)+2 ln⁡(1)=3 ln⁡(3)-5ln⁡(2)

10. Find the area inside integral f(x)=\(\frac{sec^4⁡(x)}{\sqrt{tan⁡(x)}}\) from x = 0 to π.
a) π
b) 0
c) 1
d) 2
View Answer

Answer: b
Explanation:
Given,F(x)=\(\int \frac{sec^4⁡ (x)}{\sqrt{tan⁡(x)}} dx\)
F(x)=\(\int \frac{sec^2⁡ (x) sec^2⁡ (x)}{\sqrt{tan⁡(x)}} dx\)
=\(\int \frac{1+t^2}{\sqrt{t}} dt\)
=\(\int [\frac{1}{\sqrt{t}}+t^{3/2}]dt\)
=\(2\sqrt{t}+\frac{2}{5} t^{5/2}\)
F(x)=\(\frac{2}{5} \sqrt{tan⁡(x)} [5+tan^2⁡(x)]\)
Now area inside a function f(x) from x=0 to π, is
F(π)-F(0)=0-0=0

11. Find the area inside function \(\frac{(2x^3+5x^2-4)}{x^2}\) from x = 1 to a.
a) ^a²⁄₂ + 5a – 4ln(a)
b) ^a²⁄₂ + 5a – 4ln(a) – ¹¹⁄₂
c) ^a²⁄₂ + 4ln(a) – ¹¹⁄₂
d) ^a²⁄₂ + 5a – ¹¹⁄₂
View Answer

Answer: b
Explanation: Add constant automatically
Given,
f(x) = \(\frac{(2x^3+5x^2-4)}{x^2}\),
Integrating it we get, F(x) = ^x²⁄₂ + 5x – 4ln⁡(x)
Hence, area under, x = 1 to a, is
F(a) – F(1)=^a²⁄₂ + 5a – 4ln(a) – 1/2 – 5=^a²⁄₂ + 5a – 4ln(a) – ¹¹⁄₂

12. Find the value of ∫(x⁴ – 5x² – 6x)⁴ 4x³ – 10x – 6 dx.
a) \(\frac{(x^4-5x^2-6x)^4}{4}\)
b) \(\frac{(x^4-5x^2-6x)^5}{5}\)
c) \(\frac{(4x^3-10x-6)^5}{5}\)
d) \(\frac{(4x^3-10x-6)^4}{4}\)
View Answer

Answer: b
Explanation: Add constant automatically
Given, \(\int (x^4-5x^2-6x)^4 4x^3-10x-6 dx\)
putting, \(x^4-5x^2-6x=z\), we get, \(dz=4x^3-10x-6 dx\)
\(\int z^4 dz=\frac{z^5}{5}=\frac{(x^4-5x^2-6x)^5}{5}\)

13. Temperature of a rod is increased by moving x distance from origin and is given by equation T(x) = x² + 2x, where x is the distance and T(x) is change of temperature w.r.t distance. If, at x = 0, temperature is 40 C, find temperature at x=10.
a) 473 C
b) 472 C
c) 474 C
d) 475 C
View Answer

Answer: a
Explanation: Temperature at distance x is,
T = ∫T(x) dx = ∫x² + 2x dx = ^x³⁄₃ + x² + C
At x=0 given T = 40 C
C = T(x = 0) = 40 C
At x= 10,
T(x = 10) = ¹⁰⁰⁰⁄₃ + 100 + 43 = 473 C.

14. Find the value of \(\int \frac{1}{16x^2+16x+10}dx\).
a) ¹⁄₈ sin^-1(x + ¹⁄₂)
b) ¹⁄₈ tan^-1(x + ¹⁄₂)
c) ¹⁄₈ sec^-1(x + ¹⁄₂)
d) ¹⁄₄ cos^-1(x + ¹⁄₂)
View Answer

Answer: b
Explanation: Add constant automatically
Given, \(\int \frac{1}{16x^2+16x+10}dx=\frac{1}{2}\int \frac{1}{4x^2+4x+5}dx\)
=\(\int \frac{1}{8(x^2+x+\frac{5}{4}+\frac{1}{4}+\frac{1}{4})}dx=\int \frac{1}{8[(x+\frac{1}{2})^2+1^2]}dx=\frac{1}{8}tan^{-1}(x+\frac{1}{2})\)

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