Question

Let $\alpha$ be the area of the larger region bounded by the curve $y^2=8 x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant. Then the value of $3 \alpha$ is equal to _____

Answer 1

$y=x$
& $y^2=8 x$
Solving it
$ x^2=8 x $
$ \therefore x=0,8$
$ \therefore y=0,8 $
$ x=2 \text { will intersect occur at } $
$ y^2=16 \Rightarrow y=\pm 4 $
$ \therefore \text { Area of shaded }$
$ =\int\limits_2^8(\sqrt{8 x}-x) d x=\int\limits_2^8(2 \sqrt{2} \sqrt{x}-x) d x $
$=\left[2 \sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}-\frac{x^2}{2}\right]_0^8$
$ =\left(\frac{4 \sqrt{2}}{3} \cdot 2^{9 / 2}-32\right)-\left(\frac{4 \sqrt{2}}{3} \cdot 2^{9 / 2}-2\right) $
$ =\frac{128}{3}-32-\frac{16}{3}+2=\frac{112-90}{3}=\frac{22}{3}= A$
$ \therefore 3 A =22 $