QJEEAdv08Ma402e: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma402e Solution: The area of the region bounded by the curve the -axis, and the…

Question

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The area of the region bounded by the curve $y=f(x),$ the
$x$ -axis, and the lines $x=a$ and $x=b$ , where $- \infty < a < b < - 2,$ is :

A $\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$
B $- \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$
C $\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$
D $- \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$

Accepted Answer

Required area $\int\limits_a^b {ydx = \int\limits_a^b {f(x)dx} }$

$= [f(x).x]_a^b - \int\limits_a^b {f'(x)xdx}$

$= bf(b) - af(a) - \int\limits_a^b {f'(x)xdx}$

$= bf(b) - af(a) + \int\limits_a^b {{{xdx} \over {3[{{\{ f(x)\} }^2} - 1]}}}$

$\because$ [ $f'(x) = {{dy} \over {dx}} = {{ - 1} \over {3({y^2} - 1)}} = {{ - 1} \over {3[{{\{ f(x)\} }^2} - 1]}}$ ]

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