First slide

Evaluation of definite integrals

Question

$If y = f (x) and y = g (x) are symmetrical about the line x = \frac{α + β}{2}, then$ $\int_{α}^{β} f (x) g^{'} (x) d x is equal to$

Difficult

A

$\int_{α}^{β} f^{'} (x) g (x) d x$
B

$- \int_{α}^{β} f^{'} (x) g (x) d x$
C

$\frac{1}{2} \int_{α}^{β} (f (x) g^{'} (x) - f^{'} (x) g (x)) d x$
D

$\frac{1}{2} \int_{α}^{β} (f (x) g^{'} (x) + f^{'} (x) g (x)) d x$

Solution

$f (x) i s s y m m e t r i c a b o u t x = \frac{α + β}{2}$
$\begin{array}{l} \Rightarrow f (x) = f (α + β - x) \\ \Rightarrow f (α) = f (α + β - α) = f (β) and g (α) = g (β) \\ I = \int_{α}^{β} f (x) g^{'} (x) d x = {f (x) g (x)|}_{β}^{α} - \int_{α}^{β} f^{'} (x) g (x) d x = 0 - \int_{α}^{β} f^{'} (x) g (x) d x \\ 2 I = \int_{α}^{β} f (x) g^{'} (x) - \int_{α}^{β} f^{'} (x) g (x) d x \\ I = \frac{1}{2} \int_{α}^{β} (f (x) g^{'} (x) - f^{'} (x) g (x)) d x \end{array}$

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