250+ TOP MCQs on Improper Integrals and Answers

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Engineering Mathematics Multiple Choice Questions 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)
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))ex is?
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x) + Cos(x))
Answer: b
Explanation: Add constant automatically
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)
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})
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)

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)

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)}]^3right ])
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)}]^3right ])
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)}]^3right ])
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)}]^3right ])
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=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)}]^3right])

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
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 = aeb to a.
a) b2⁄2
b) b⁄2
c) b
d) 1
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)
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)

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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
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)
=(2sqrt{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
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})
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
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.

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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)
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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