Question

Find the area of region bounded by the curve
$y=x^2,y=x+6$ and $x=0$.

Answer 1

REF.Image.

$x^2=x+6$

$(x-3)(x+2)=0$

$x=3$ or $-2$

so area

${\int }_{-2}^3\left(\right(x+6)-x^2)dx$

$\Rightarrow {[\frac{x^2}{2}+6x-\frac{x^3}{3}]}_{-2}^3$

$\Rightarrow (\frac{9}{2}+18-\frac{27}{3})-(\frac{4}{2}-12-\frac{(-8)}{3})$

$=\frac{9}{2}+18-\frac{27}{3}-\frac{4}{2}+12-\frac{8}{3}$

$\frac{5}{2}+30-\frac{35}{3}$

$\frac{15+180-70}{6}=\frac{125}{6}$