Question

The area enclosed by the curve $y=x^2+2x-3$ and $x$ axis is

• A. $\frac{32}{3}$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $\frac{64}{3}$

Answer 1

The correct option is A $\frac{32}{3}$


$y=x^2+2x-3\Rightarrow y=(x+3)(x-1)$
The area $=\underset{-3}{\overset{1}{\int }}{x}^2+2x-3dx={[\frac{x^3}{3}+x^2-3x]}_{-3}^1=[\frac{28}{3}-8-12]=-\frac{32}{3}$
As area cannot be negative
So Area enclosed $=\frac{32}{3}$ sq. units