Exercise
Find the area of the region bounded by the graphs of the equations:
y=x^2, y=-x+6, y=0
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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