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Question: If \(\int_{0}^{x}{f(t)}dt = x + \int_{x}^{1}{tf(t)}dt,\) then the value of f(1) is...

If ∫0xf(t)dt=x+∫x1tf(t)dt,\int_{0}^{x}{f(t)}dt = x + \int_{x}^{1}{tf(t)}dt, then the value of f(1) is

A

12\frac{1}{2}

B

0

C

1

D

−12- \frac{1}{2}

Answer

12\frac{1}{2}

Explanation

Solution

∫0xf(t)dt=x+∫x1tf(t)dt,\int_{0}^{x}{f(t)dt} = x + \int_{x}^{1}{tf(t)dt,}

⇒ ddx(∫0xf(t)dt)=ddx(x+∫x1tf(t)dt)\frac{d}{dx}\left( \int_{0}^{x}{f(t)dt} \right) = \frac{d}{dx}\left( x + \int_{x}^{1}{tf(t)dt} \right)

⇒ f(x)=1+0−xf(x)f(x) = 1 + 0 - xf(x)

[Using Leibnitz's Rule]

⇒ f(x)=1−xf(x)f(x) = 1 - xf(x)

⇒ f(x)=1x+1⇒f(1)=12f(x) = \frac{1}{x + 1} \Rightarrow f(1) = \frac{1}{2}