Definite Integral – Finding area between a polynomial and a line – Exercise 7002

Exercise

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

$y=-x^2+5x+6, y=-x+6$

Final Answer

Show final answer

Solution

First, we find out how the area looks like:

$-x^2+5x+6=-x+6$

$x^2-6x=0$

$x(x-6)=0$

$x=0, x=6$

The area looks like this:

$S=\int_0^6 -x^2+5x+6-(-x+6) dx=$

$=\int_0^6 -x^2+6x dx=$

$=[-\frac{x^3}{3}+6\cdot\frac{x^2}{2}]_0^6=$

$=[-\frac{x^3}{3}+3x^2]_0^6=$

$=-\frac{6^3}{3}+3\cdot 6^2-(-\frac{0^3}{3}+3\cdot 0^2)=$

$=-\frac{216}{3}+3\cdot 36-0=$

$=36$

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