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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


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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