# Engineering Mathematics Questions and Answers – Improper Integrals – 2

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) – 12 ln⁡(1 + x2)
b) xtan-1 (x) – 12 ln⁡(1 + x2)
c) xsec-1 (x) – 12 ln⁡(1 + x2)
d) tan-1 (x) – 12 ln⁡(1 + x2)

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) – 12 ln⁡(1 + x2).

2. Integration of (Sin(x) + Cos(x))ex is?
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x) + Cos(x))

Let f(x) = ex Sin(x)
∫ex Sin(x)dx = ex Sin(x) – ∫ex Cos(x)dx
∫ex Sin(x)dx + ∫ex Cos(x)dx = ∫ex [Cos(x) + Sin(x)]dx = ex 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)

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) $$\sum_{i=0}^n(-1)^i u_i v^{n-i}$$

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)

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)

6. Find the value of ∫x3 ex e2x e3x….enx 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)$$\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]$$

By, f(x)=$$\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}$$
Let, u = x3 and v=ex e2x e3x…..enx=ex(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) a22 + aSin(a) + Cos(a) – 1
b) a33 + aSin(a) + Cos(a)
c) a33 + aSin(a) + Cos(a) – 1
d) a33 + Cos(a) + Sin(a) – 1

Explanation: Given, f(x) = x2 + xCos(x)
Hence, F(x) = ∫x2 + xCos(x) dx = x33 + xSin(x) + Cos(x)
Hence, area inside f(x) is,
F(a) – F(0) = a33 + aSin(a) + Cos(a) – 1.

8. Find the area ln(x)x from x = x = aeb to a.
a) b22
b) b2
c) b
d) 1

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)

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

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) a22 + 5a – 4ln(a)
b) a22 + 5a – 4ln(a) – 112
c) a22 + 4ln(a) – 112
d) a22 + 5a – 112

Given,
f(x) = $$\frac{(2x^3+5x^2-4)}{x^2}$$,
Integrating it we get, F(x) = x22 + 5x – 4ln⁡(x)
Hence, area under, x = 1 to a, is
F(a) – F(1)=a22 + 5a – 4ln(a) – 1/2 – 5=a22 + 5a – 4ln(a) – 112

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) $$\frac{(4x^3-10x-6)^4}{4}$$

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

Explanation: Temperature at distance x is,
T = ∫T(x) dx = ∫x2 + 2x dx = x33 + x2 + C
At x=0 given T = 40 C
C = T(x = 0) = 40 C
At x= 10,
T(x = 10) = 10003 + 100 + 43 = 473 C.

14. Find the value of $$\int \frac{1}{16x^2+16x+10}dx$$.
a) 18 sin-1(x + 12)
b) 18 tan-1(x + 12)
c) 18 sec-1(x + 12)
d) 14 cos-1(x + 12)

Given, $$\int \frac{1}{16x^2+16x+10}dx=\frac{1}{2}\int \frac{1}{4x^2+4x+5}dx$$
=$$\int \frac{1}{8(x^2+x+\frac{5}{4}+\frac{1}{4}+\frac{1}{4})}dx=\int \frac{1}{8[(x+\frac{1}{2})^2+1^2]}dx=\frac{1}{8}tan^{-1}(x+\frac{1}{2})$$

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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