First slide

Evaluation of definite integrals

Question

Let $I_{1} = \int_{a}^{b} (\int_{a}^{x} f (t) d t) d x$ and $I_{2} = \int_{a}^{b} (b - x) f (x) d x$ then

Moderate

A

$I_{1} = I_{2}$
B

$I_{1} = (b - a) + I_{2}$
C

$I_{1} = (b - a) I_{2}$
D

none of these

Solution

Let $F (x) = \int_{α}^{x} f (t) d t$ then $F^{'} (x) = f (x)$
$\begin{array}{l} I_{1} = \int_{a}^{b} F (x) d x = {x F (x)|}_{a}^{b} - \int_{a}^{b} x f (x) d x \\ = b F (b) - a F (a) - \int_{a}^{b} x f (x) d x \\ = \int_{a}^{b} (b - x) f (x) d x = I_{2} . \end{array}$

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