# Definite Integral – Finding area between a polynomial and a line – Exercise 7006

Exercise

Find the area of the region bounded by the graphs of the equations:

$$y=x^2, y=2x+3$$

$$S=10\frac{2}{3}$$

Solution

First, we find out how the area looks like:

$$x^2=2x+3$$

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

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

$$x=-1, x=3$$

The area looks like this:

$$S=\int_{-1}^3 2x+3-x^2 dx=$$

$$=[2\cdot\frac{x^2}{2}+3x-\frac{x^3}{3}]_{-1}^3=$$

$$=[x^2+3x-\frac{x^3}{3}]_{-1}^3=$$

$$=3^2+3\cdot 3-\frac{3^3}{3}-({(-1)}^2+3\cdot (-1)-\frac{{(-1)}^3}{3})=$$

$$=9+9-9-(1-3+\frac{1}{3})=$$

$$=9-1+3-\frac{1}{3}=$$

$$=10\frac{2}{3}$$

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