# Engineering Mathematics Questions and Answers – Improper Integrals – 1

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Improper Integrals – 1”.

1. Integration of function is same as the ___________
a) Joining many small entities to create a large entity
b) Indefinitely small difference of a function
c) Multiplication of two function with very small change in value
d) Point where function neither have maximum value nor minimum value

Explanation: Integration of function is same as the Joining many small entities to create a large entity.

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))

Explanation: 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. Integration of (Sin(x) – Cos(x))ex is ___________
a) -ex Cos(x)
b) ex Cos(x)
c) -ex Sin(x)
d) ex Sin(x)

Let f(x) = ex Sin(x)
∫ ex Sin(x)dx = -ex Cos(x) + ∫ ex Cos(x)dx
∫ ex Sin(x)d-∫ ex Cos(x)dx = ∫ ex [Sin(x)-Cos(x)]dx = -ex Cos(x).

4. Value of ∫ Cos2 (x) Sin2 (x)dx.
a) $$\frac{1}{8} [x-\frac{Cos(2x)}{2}]$$
b) $$\frac{1}{4} [x-\frac{Cos(2x)}{2}]$$
c) $$\frac{1}{8} [x-\frac{Sin(2x)}{2}]$$
d) $$\frac{1}{4} [x-\frac{Sin(2x)}{2}]$$

Given,f(x)=$$\int Cos^2 (x) Sin^2 (x)dx=\frac{1}{4} \int Sin^2 (2x) dx=\frac{1}{4} \int \frac{[1-Cos(2x)]}{2} dx=\frac{1}{8} [x-\frac{Sin(2x)}{2}]$$

5. If differentiation of any function is zero at any point and constant at other points then it means?
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point

Explanation: Since slope of a function is given by dydx at that point. Hence, when dydx = 0 means slope of a function is zero i.e, parallel to x axis.
Function is not a constant function since it has finite value at other points.
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6. If differentiation of any function is infinite at any point and constant at other points then it means ___________
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point

Explanation: Since slope of a function is given by dydx at that point.Hence,when dydx = ∞ means slope of a function is 90 degree i.e,parallel to y axis.

7. Integration of function y = f(x) from limit x1 < x < x2 , y1 < y < y2, gives ___________
a) Area of f(x) within x1 < x < x2
b) Volume of f(x) within x1 < x < x2
c) Slope of f(x) within x1 < x < x2
d) Maximum value of f(x) within x1 < x < x2

Explanation: Integration of function y=f(x) from limit x1 < x < x2 , y1 < y < y2, gives area of f(x) within x1 < x < x2.

8. Find the value of ∫ ln⁡(x)x dx.
a) 3a2
b) a2
c) a
d) 1

Given, f(x)=$$\int \frac{ln⁡(x)}{x} dx$$
Let, z=ln⁡(x)=>dz=$$\frac{dx}{x}=>f(x)=\int zdz=z^2/2=\frac{ln^2⁡(x)}{2}$$

9. Find the value of ∫t(t+3)(t+2) dt, is?
a) 2 ln⁡(t+3)-3 ln⁡(t+2)
b) 2 ln⁡(t+3)+3 ln⁡(t+2)
c) 3 ln⁡(t+3)-2 ln⁡(t+2)
d) 3 ln⁡(t+3)+2ln⁡(t+2)

Given, et = x => dx = et dt,
Given, f(x)=$$\int \frac{ln⁡(x)}{x} dx$$
Let, z=ln⁡(x)=>dz=$$\frac{dx}{x}=>f(x)=\int zdz=\frac{z^2}{2}=\frac{ln^2⁡(x)}{2}$$

10. Find the value of ∫ cot3(x) cosec4 (x).
a) –$$[\frac{cot^4⁡(x)}{4}+\frac{cosec^6⁡(x)}{6}]$$
b) –$$[\frac{cosec^4⁡(x)}{4}+\frac{cosec^6⁡(x)}{6}]$$
c) –$$[\frac{cot^4⁡(x)}{4}+\frac{cot^6⁡(x)}{6}]$$
d) –$$[\frac{cosec^4⁡(x)}{4}+\frac{cot^6⁡(x)}{6}]$$

Given, $$\int cot^3⁡(x)cosec^4 (x)dx=-\int cot^3⁡(x)cosec^2 (x)dcot(x)$$
=-$$\int t^3 (1+t^2)dt=-[\frac{t^4}{4}+\frac{t^6}{6}]=-[\frac{cot^4⁡(x)}{4}+\frac{cot^6⁡(x)}{6}]$$

11. Find the value of $$\int \frac{sec^4⁡(x)}{\sqrt{tan⁡(x)}} dx$$.
a) $$\frac{2}{5}\sqrt{tan⁡(x)}[5+sec^2⁡(x)]$$
b) $$\frac{2}{5}\sqrt{sec⁡(x)}[5+tan^2⁡(x)]$$
c) $$\frac{2}{5}\sqrt{tan⁡(x)}[6+tan^2⁡(x)]$$
d) $$\frac{2}{5}\sqrt{tan⁡(x)}[5+tan^2⁡(x)]$$

Given, $$\int \frac{sec^4⁡(x)}{\sqrt{tan⁡(x)}} dx$$
=$$\int \frac{sec^2⁡(x) sec^2⁡(x)}{\sqrt{tan⁡(x)}} dx$$
=$$\int \frac{1+t^2}{\sqrt{t}} dt$$
=$$\int [\frac{1}{\sqrt{t}}+t^{3/2}]dt$$
=$$2\sqrt{t}+\frac{2}{5} t^{5/2}$$
=$$\frac{2}{5}\sqrt{tan⁡(x)}[5+tan^2⁡(x)]$$

12. Find the value of $$\int \frac{1}{4x^2+4x+5} dx$$.
a) 18 sin(-1)⁡(x + 12)
b)14 tan(-1)⁡(x + 12)
c) 18 sec(-1)⁡(x + 12)
d) 14 cos(-1)⁡(x + 12)

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

13. Find the value of $$\int \sqrt{4x^2+4x+5} dx$$.
a) $$2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{{(x+\frac{1}{2})^2+1)}}\right ]+ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]$$
b) $$2\left [\frac{1}{2} \sqrt{(x+\frac{1}{2})^2+1)}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]$$
c) $$2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{(x+\frac{1}{2})^2+1)}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]$$
d) $$2\left [(x+\frac{1}{2}) \sqrt{{(x+\frac{1}{2})^2+1)}}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]$$

Given, $$\int \sqrt{4x^2+4x+5} dx=\int 2\sqrt{(x+\frac{1}{2})^2+1^2} dx$$
=$$\int 2\sqrt{t^2+1^2} dt=2\left [\frac{1}{2} t\sqrt{t^2+1}\right ]+\frac{1}{2} ln⁡[t+\sqrt{t^2+1}]$$
=$$2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{(x+\frac{1}{2})^2+1)} \right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+1/2)^2+1}\right ]$$

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