Question
Evaluate: $\int_3^8 \frac{(11-x)^2}{x^2+(11-x)^2} d x$
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Answer
$ \text { Let } I =\int_3^8 \frac{(11-x)^2}{x^2+(11-x)^2} d x \quad \ldots \ldots . . \text { (i) }$
$=\int_3^8 \frac{[11-(1-x)]^2}{(11-x)^2+[11-(11-x)] 2} d x \quad \ldots \ldots . .\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore I =\int_3^8 \frac{x^2}{(11-x)^2+x^2} d x \ldots \ldots \text { (ii) } $
Adding (i) and (ii), we get
$ 2 I =\int_3^8 \frac{(11-x)^2}{x^2+(11-x)^2} d x+\int_3^8 \frac{x^2}{(11-x)^2+x^2} d x$
$=\int_3^8 \frac{(11-x)^2+x^2}{x^2+(11-x)^2} d x$
$\therefore 2 I =\int_3^8 1 \cdot d x$
$\therefore I =\frac{1}{2}[x]_3^8$
$\therefore I =\frac{1}{2}(8-3)$
$\therefore I =\frac{5}{2} $
$=\int_3^8 \frac{[11-(1-x)]^2}{(11-x)^2+[11-(11-x)] 2} d x \quad \ldots \ldots . .\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore I =\int_3^8 \frac{x^2}{(11-x)^2+x^2} d x \ldots \ldots \text { (ii) } $
Adding (i) and (ii), we get
$ 2 I =\int_3^8 \frac{(11-x)^2}{x^2+(11-x)^2} d x+\int_3^8 \frac{x^2}{(11-x)^2+x^2} d x$
$=\int_3^8 \frac{(11-x)^2+x^2}{x^2+(11-x)^2} d x$
$\therefore 2 I =\int_3^8 1 \cdot d x$
$\therefore I =\frac{1}{2}[x]_3^8$
$\therefore I =\frac{1}{2}(8-3)$
$\therefore I =\frac{5}{2} $
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