First slide

Evaluation of definite integrals

Question

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

Moderate

A

$\frac{1}{2}$
B

0
C

1
D

$- \frac{1}{2}$

Solution

We have,
$\begin{array}{l} \int_{0}^{x} f (t) dt = x + \int_{x}^{1} tf (t) dt \\ \Rightarrow f (x) = 1 + 0 - xf (x) [using Leibnitz's rule) \\ \Rightarrow f (x) = \frac{1}{1 + x} \\ f (1) = \frac{1}{2} \end{array}$

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