Integrals- Quiz 4

 

Integrals Quiz-4

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Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

Q1. 

 

The primitive of the function x |cos⁡x| when π/2< x<π is given by

•  cos⁡x+x sin⁡x+C
•  -cos⁡x-x sin⁡x+C
•  xsin x-xcosx+C
•  None of these

Solution
f(x)=x|cos⁡x |,π/2< x<π=-x cos⁡x,because cos⁡x is negative in (π/2,π) ∴ the required primitive function = ∫-x cos⁡xdx Now, use integration by parts

Q2.Let g(x)=

where f is such that 1/2≤f(t)≤1, for t∈[0,1] and 0≤f(t)≤1/2, for t∈[1,2]. Then g(2) satisfies the inequality

•  -3/2≤g(2)<1/2
•  1/2≤g(2)<3/2
•  3/2≤g(2)<5/2
•  2≤g(2)<4

Solution

 

 

Q3.

 

Let f:R→R be a continuous function and f(x)=f(2x) is true ∀ x∈R. If f(1)=3, then the value of

 

is equal to

•  6
  

•  
  0
  
  
  
  

•  3f(3)
  

•  2f(0)
  

Solution
f(2x)=f(x)=f(x/2)=f(x/2^2 )=⋯=f(x/2^n ) So, when n→∞⇒f(2x)=f(0) (f(x) is continuous) i.e., f(x) is a constant function ⇒f(x)=f(1)= 3,∫_(-1)^1f(f(x) )dx=∫_(-1)^13dx=6

Q4. The value of the definite integral

 

•  1
•  1+( sin⁡1)
•  1-( sin⁡1)
•   (sin⁡1)-1

Solution

Q5.

 

If ∫dx/(x^2 (x^n+1)^((n-1)/n) )=-[f(x)]^(1/n)+c,then f(x) is

•  (1+x^n )
•  1+x^(-n)
•  x^n+x^(-n)
•  None of these

Solution
 We have ∫dx/(x^2 (x^n+1)^((n-1)/n) ) =∫dx/(x^2 x^(n-1) (1+1/x^n )^((n-1)/n) ) =∫dx/(x^(n+1) (1+x^(-n) )^((n-1)/n) ) Put 1+x^(-n)=t ∴ nx^(-n-1) dx=dt⇒dx/x^(n+1) =-dt/n ⇒∫dx/(x^2 (x^n+1)^((n-1)/n) )=-1/n ∫dt/t^((n-1)/n) =-1/n ∫t^(1/n-1) dt=-1/n t^(1/n-1+1)/(1/n-1+1)+C =-t^(1/n)+C=-(1+x^(-n))^(1/n)+C

Q6. The value of

 

is

•  (b-a)^4/6^4
•  (b-a)^8/280
• (b-a)^7/7^3
•  None of these

Solution

Q7.  If f(π)=2 and

 

(f(x)+ f^'' (x))sin⁡x dx=5, then f(0) is equal to (it is given that f(x) is continuous in [0,π])

•  7
•  3
•  5
•  1

Solution

 

Q8. If

 

is equal to

•  0
•  1
•  2
•  3

Solution

 

Q9.If xf(x)=3f^2 (x)+2,then∫(2x^2-12xf(x)+f(x))/((6f(x)-x) (x^2-f(x))^2 ) dx equals

•  1/(x^2-f(x))+c
•  1/(x^2+f(x))+c
•  1/(x-f(x))+c
•  1/(x+f(x))+c

Solution
f^' (x)=(f(x))/(6f(x)-x) Now I=∫(2x(x-6f(x) )+f(x))/((6f(x)-x)(x^2-f(x))^2 )dx⇒I=-∫(2x-f^' (x))/(x^2-f(x))^2 dx=1/(x^2-f(x))+C

Q10. The value of the definite integral

 

•  0
•  π/2
•  π
• 2π

Solution