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• Notes

Find the area of the region bounded by
$y=e^{x} \text { bounded below by } y=x \text { and bounded on the sides by } x=0 \text { and } x=1$

$f(x)=e^{x} \quad g(x)=x$

$d x : 0 \rightarrow 1 \quad a=0 \quad b=1$

$=\int_{0}^{1}\left(e^{x}-x\right) d x=\left[e^{x}-\frac{x^{2}}{2}\right]^{1}$

$=\left[e^{(1)}-\frac{(1)^{2}}{2}\right]-\left[e^{0}-0\right]=e-\frac{1}{2}-1=e-1.5$

Find the area of the region enclosed by parabolas

$y=x^{2} \text { and } y=2 x-x^{2}$

Inter section points $\rightarrow x^{2}=2 x-x^{2} \rightarrow x^{2}+x^{2}-2 x=0 \rightarrow 2 x^{2}-2 x=0$

$\frac{2 x^{2}}{2}-\frac{2 x=0}{2}$

$x^{2}-x=0 \rightarrow x(x-1)=0$

إما

$x=0$

أو

$x-1=0 \longrightarrow x=1$

$y=0 \quad y=1$

$(0,0)(1,1)$

Area
$=\int_{0}^{1}[f(x)-g(x)] d x$

$=\int_{0}^{1}\left(2 x-x^{2}-x^{2}\right) d x=\int_{0}^{1}\left(2 x-2 x^{2}\right) d x$

$=\left[\frac{2 x^{2}}{x}-\frac{2 x^{3}}{3}\right]_{0}^{1}=\left[x^{2}-\frac{2 x^{3}}{3}\right]_{0}^{1}=\left((1)^{2}-\frac{2 (1)^{3}}{3}\right)-(0)$

$=1-\frac{2}{3}=\frac{1}{3}$