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:

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

Final Answer

Show final answer

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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