Engineering Mathematics Multiple Choice Questions 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
Answer: a
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))
Answer: b
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)
Answer: a
Explanation: Add constant automatically
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}])
Answer: c
Explanation: Add constant automatically
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
Answer: a
Explanation: Since slope of a function is given by dy⁄dx at that point. Hence, when dy⁄dx = 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.
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
Answer: a
Explanation: Since slope of a function is given by dy⁄dx at that point.Hence,when dy⁄dx = ∞ 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
Answer: a
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
Answer: a
Explanation: Add constant automatically
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)
Answer: c
Explanation: Add constant automatically
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}])
Answer: c
Explanation: Add constant automatically
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)])
Answer: d
Explanation: Add constant automatically
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)
=(2sqrt{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) 1⁄8 sin(-1)(x + 1⁄2)
b)1⁄4 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}{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) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+lnleft [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
b) (2left [frac{1}{2} sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} lnleft [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
c) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} lnleft [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
d) (2left [(x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+frac{1}{2} lnleft [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
Answer: c
Explanation: Add constant automatically
Given, (int sqrt{4x^2+4x+5} dx=int 2sqrt{(x+frac{1}{2})^2+1^2} dx)
=(int 2sqrt{t^2+1^2} dt=2left [frac{1}{2} tsqrt{t^2+1}right ]+frac{1}{2} ln[t+sqrt{t^2+1}])
=(2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)} right ]+frac{1}{2} lnleft [(x+frac{1}{2})+sqrt{(x+1/2)^2+1}right ])