Question

Let f(x) = ₀ˣ∫ g(t)dt, were g is a non zero even function. If f(x+5) = g(x) , then ₀ˣ∫ f(t)dt
equals (A) ⁵∫ₓ₊₅ g(t)dt
(B) 2∫₅ˣ⁻⁵ g(t)dt
(C) ˣ⁺⁵∫₅ g(t)dt
(D) 5∫⁵ₓ₊₅ g(t)dt

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Therefore

$\int^x_0f(x) dx=\int^5_{5+x} g(t)dt$

Step-by-step explanation:

Given,

$f(x)= \int^x_0 g(t) dt$

Putting x=-x

$f(-x)= \int^{-x}_0 g(t) dt$

Putting t = -u, dt = -du

$=\int^u_0g(-u) (-du)$  

$=-\int^u_0g(-u) du$

$=-\int^u_0g(u) du$     [ since g is a non zero even function]

= - f(x)

Therefore f is a odd function.

And

$f(x+5)= g(x)$

Putting x= -x

$f(-x+5)= g(-x)$

                = g(x)      [ since g is a non zero even function]

     f(-x+5)=f(x+5)

Let

$I=\int^x_0f(x) dx$

Putting x= 5+u , dx= du,lower limit x=0, u+5=0⇒u= -5 and upper limit x=x , u= x-5

   $=\int^{x-5}_{-5}f(5+u) du$

  $=\int ^{x-5}_{-5}g(u) du$

  $=\int ^{x-5}_{-5}f'(u) du$

  $=f(x-5)-f(-5)$

  $=-f(-x+5)+f(5)$    [ since f is odd function f(-x)=-f(x)]

  $=f(5)-f(-x+5)$

 $=f(5)-f(x+5)$   [ ∵ f(-x+5)=f(5+x)  from (i)]

  $=\int^5_{5+x} f'(t)dt$

  $=\int^5_{5+x} g(t)dt$

 Therefore

$\int^x_0f(x) dx=\int^5_{5+x} g(t)dt$