First slide

Evaluation of definite integrals

Question

Let $f_{1} (x) = \int_{0}^{x} f (t) d t, f_{2} (x) = \int_{0}^{x} f_{1} (t) d t$ and $f_{3} (x)$
$= \int_{0}^{x} f_{2} (t) d t$ if $f_{3} (x) = A \int_{0}^{x} f (t) (x - t)^{2} d t$ then the value of $A$ is

Difficult

A

1
B

$1 / 2$
C

2
D

none of these

Solution

Note that $\frac{d}{d x} f_{3} (x) = f_{2} (x), \frac{d}{d x} f_{2} (x) = f (x)$ and $\frac{d}{d x} f_{1} (x) = f (x)$ . Integrating by parts
$\begin{array}{l} f_{3} (x) = {t f_{2} (t)|}_{0}^{x} - \int_{0}^{x} t f_{1} (t) d t \\ = x f_{2} (x) - \int_{0}^{x} t f_{1} (t) d t = \int_{0}^{x} (x - t) f_{1} (t) d t \\ = - {\frac{(x - t)^{2}}{2} f_{1} (t)|}_{0}^{x} + \frac{1}{2} \int_{0}^{x} (x - t)^{2} f (t) d t \end{array}$
$= \frac{1}{2} \int_{0}^{x} (x - t)^{2} f (t) d t .$ Thus $A = \frac{1}{2} .$

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