Answer to Question #342419 in Calculus for Jane

Question

Find the area of f(x) = 3+2x-x² above the x-axis.

Answer 1

$3+2x-x^2=0$

$-(x+1)(x-3)=0$

$x_1=-1, x_2=3$ $A=\displaystyle\int_{-1}^{3}(3+2x-x^2)dx$

$=\bigg[3x+x^2-\dfrac{x^3}{3}\bigg]\begin{matrix} 3 \\ -1 \end{matrix}$

$=9+9-9-(-3+1+\dfrac{1}{3})=\dfrac{32}{3}({units}^2)$