Question

Let $\alpha$ be the area of the larger region bounded by the curve $y^2=8x$ 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

Answer: 22

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