First slide

Evaluation of definite integrals

Question

If $f (x) = (1 + \tan x) (1 + \tan (π / 4 - x))$ and $g (x)$ is a function with domain R, then
$\int_{0}^{1} x^{3} g \circ f (x) d x$ is

Moderate

A

$\frac{1}{2} g (π / 4)$
B

$\frac{1}{4} g (2)$
C

$\frac{1}{4} g (1)$
D

none of these

Solution

$\begin{matrix} f (x) = (1 + \tan x) (1 + \frac{1 - \tan x}{1 + \tan x}) \\ = (1 + \tan x) (\frac{2}{1 + \tan x}) = 2 . \end{matrix}$
So
$\begin{array}{l} \int_{0}^{1} x^{3} g \circ f (x) d x = \int_{0}^{1} x^{3} g (2) d x \\ = {\frac{x^{4} g (2)}{4}|}_{0}^{1} = \frac{1}{4} g (2) . \end{array}$

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