Question

Using integration find the area of the region bounded between the line $x=4$ and the parabola $y^2=16x$.

Answer 1

$y^2=16x$

$y=4\sqrt{x}$

Total area $=p+p$

$=2p$

$=2{\int }_0^{4}4x^{\frac{1}{2}}dx$

$=8{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^4$

$=\frac{8\times 2}{3}\left[\right(4{)}^{\frac{3}{2}}-0]$

$=\frac{16}{3}\left[\right(2^2{)}^{\frac{3}{2}}]=\frac{64}{3}units$.