# Definite Integral – Finding area between parabola, line and axis-x – Exercise 7024

Exercise

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

$$y=x^2, y=-x+6, y=0$$

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

Solution

First, we find out how the area looks like:

$$x^2=-x+6$$

$$x^2+x-6=0$$

$$(x-2)(x+3)=0$$

$$x=2, x=-3$$

The area looks like this:

Hence, it is a sum of two disjoint areas:

We will calculate them seperately:

$$S=S_1+S_2$$

The first area:

$$S_1=\int_0^2 x^2 dx=$$

$$=[\frac{x^3}{3}]_0^2=$$

$$=\frac{2^3}{3}-\frac{0^3}{3}=$$

$$=\frac{8}{3}$$

The second area:

$$S_2=\int_2^6 -x+6 dx=$$

$$= [-\frac{x^2}{2}+6x]_2^6=$$

$$=-\frac{6^2}{2}+6\cdot 6-(-\frac{2^2}{2}+6\cdot 2)=$$

$$=-\frac{36}{2}+36-(-2+12)=$$

$$=18-10=$$

$$=8$$

Hence, we got

$$S_2=8$$

Lastly, we sum up the results:

$$S=S_1+S_2=$$

$$=\frac{8}{3}+8=$$

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

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