Question

Find the area between the curves $y=x^2-2x-3$, $x$-axis and the lines $x=-3$ and $x=5$.

Answer 1

The given curve $y=x^2-2x-3$
The curve meet $x$ axis
Put $y=0$
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\Rightarrow x=3,-1$
The curve lies above the $x$-axis between $x=-3$ and $x=-1$ and $x=3$ and $x=5$.
The curve lies below $x$-axis between $x=-1$ and $x=3$.
The required area $={\int }_a^{b}ydx+{\int }_c^d-ydx+{\int }_e^{f}ydx$
$={\int }_{-3}^{-1}(x^2-2x-3)dx-{\int }_{-1}^3(x^2-2x-3)dx+{\int }_3^5(x^2-2x-3)dx$
$={[\frac{x^3}{3}-x^2-3x]}_{-3}^{-1}-{[\frac{x^3}{3}-x^2-3x]}_{-1}^3+{[\frac{x^3}{3}-x^2-3x]}_3^5$
$=\frac{-1}{3}-1+3-(\frac{-27}{3}-9+9)-(\frac{+27}{3}-9-9)+(\frac{-1}{3}-1+3)+\frac{125}{3}-25-15-(\frac{-27}{3}-9-9)$
$=(\frac{-1}{3}+2)-(-9)-(-9)+(\frac{-1}{3}+2)+(\frac{125}{3}-40)-(-a)$
$=-\frac{1}{3}+2+9+9-\frac{1}{3}+2+\frac{125}{3}-40+9$
$=(-\frac{1}{3}-\frac{1}{3}+\frac{125}{3})+(-9)$
$=\frac{-1-1+125}{3}-9=\frac{123}{3}-9$
$=41-9=32$ square units.