Question

Find the area bounded by the curve $y=4-x^2$ and the lines $y=0,y=3$.

Answer 1

$\begin{array}{c}y=4-x^2,y=0,y=3\\ Area=\int ydxof\int xdx\\ ify=0,then,\\ 0=4-x^2\\ x^2=4=7x=2\\ ify=3,then,\\ 3=4-x^2\\ x^2=1\Rightarrow x=1\\ area={\int }_1^{2}ydx\\ ={\int }_1^2\left(4-x^2\right)dx\\ ={\left[4x\right]}_1^2-\frac{1}{3}{\left[x^3\right]}_1^2\\ =4\left[2-1\right]-\frac{1}{3}\left[8-1\right]\\ =4-\frac{7}{3}=\frac{5}{3}\end{array}$