Exercise
Find the area of the region bounded by the graphs of the equations:
y=x^2, y=2x+3
Final Answer
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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