Answer to Question #188470 in Calculus for Jethro

Question

Determine the area of the region bounded by y = x² and y = 2x + 3

Answer 1

Let us determine the points of intersection of $y=x^2$ and $y=2x+3$ :

$x^2 = 2x + 3, \\ x^2 - 2x-3 = 0, \\ D = (-2)^2 - 4\cdot 1\cdot (-3) = 16=4^2, \\ x_{1,2} = \dfrac{2\pm 4}{2} = 3\,\,\,\mathrm{or} \,\, -1.$

So we need to evaluate the integral equal to the difference of areas under the graphs when $x\in[-1;3]$

$S=\int\limits_{-1}^3 \left(2x+3 - x^2 \right)\,dx = \left( x^2 + 3x - \dfrac{x^3}{3}\right) \Big|_{-1}^3 \,, \\ S = \left(3^2 + 3\cdot3 - \dfrac{3^3}{3} \right) - \left( (-1)^2 + 3\cdot(-1) - \dfrac{(-1)^3}{3}\right), \\ S = \dfrac{32}{3}.$