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) – 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)
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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))
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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)
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4. Value of ∫uv dx,where u and v are function of x

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Answer: c
Explanation: Add constant automatically

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)
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Answer: a
Explanation: Add constant automatically

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 e^x e^2x e^3x……..e^nx dx

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Answer: a
Explanation: Add constant automatically

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

9. Find the area inside a function f(t) = t/(t+3)(t+2) 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)
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Answer: d
Explanation:

10. Find the area inside integral from x = 0 to π
a) π
b) 0
c) 1
d) 2
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Answer: b
Explanation:

11. Find the area inside function 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
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Answer: b
Explanation: Add constant automatically
Given,
f(x) = ,
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

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Answer: b
Explanation: Add constant automatically

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
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14. Find the value of
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)
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Answer: b
Explanation: Add constant automatically

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