Question

Draw a rough sketch of the curve $y^2=4x$ and find the area of the region enclosed by the curve and the line $y=x$.

Answer 1

$y^2=4x\Rightarrow y=2\sqrt{x}$

Area of region enclosed $=\Delta ={\int }_0^4(2\sqrt{x}-x)dx$

$\Delta =2{\int }_0^4\sqrt{x}dx-{\int }_0^{4}xdx\Delta =2{\left[\frac{x^{1/2+1}}{\frac{1}{2}+1}\right]}_0^4-{\left[\frac{x^2}{2}\right]}_0^4\Delta =\frac{4}{3}{\left[x^{3/2}\right]}_0^4-\frac{1}{2}{\left[x^2\right]}_0^4\Delta =\frac{4}{3}(8-0)-\frac{1}{2}(16-0)\Delta =\frac{32-24}{3}=\frac{8}{3}$