First slide

Evaluation of definite integrals

Question

If $y = f (x)$ be an invertible function with inverse $g$ and $h (x) = x f (x),$ then $\int_{f (a)}^{f (b)} g (x) d x + \int_{a}^{b} f (x) d x$ is equal to

Moderate

A

$h (a) - h (b)$
B

$h (b) - h (a)$
C

$h (a) + h (b)$
D

$b h (b) - a h (a)$

Solution

Put $g (x) = t \Rightarrow x = f (t) \Rightarrow d x = f^{'} (t) d t$
$\begin{array}{l} I = \int_{a}^{b} t f^{'} (t) d t + \int_{a}^{b} f (x) d x \\ = [t f (t)]_{a}^{b} - \int_{a}^{b} 1 \cdot f (t) d t + \int_{a}^{b} f (x) d x \\ = h (t)]_{a}^{b} = h (b) - h (a) \end{array}$

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