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The diagram shows the curve y=(1/2x-2)^6+5. Find the area of the shaded region.

The diagram shows the curve y=(1/2x-2)^6+5. Find the area of the shaded region.-example-1
User Thomite
by
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1 Answer

7 votes
so the line that is the limit is where the equation crosses the y axis or where x=0

so
y=(1/2(0)-2)^2+5
y=(-2)^2+5
y=4+5
y=9
at y=9

the upper bound is y=9

alrighty

we will do
let's call the curve that is 6th degree f(x)
and the y=9, g(x)

f(x)=(1/2x-2)^6+5
g(x)=9

find where they intersect

9=(1/2x-2)^6+5
4=(1/2x-2)^6
fancy math
x=8

so the area under the curve will be

\int\limits^8_0 {g(x)-f(x)} \, dx because g(x) is above f(x)
so


\int\limits^8_0 {g(x)-f(x)} \, dx= \int\limits^8_0 {9-(((1)/(2)x-2)^6+5)} \, dx=
using our calculator because I can't figure out what
\int\limits ((1)/(2)x-2)^6} \, dx is

\int\limits^8_0 {9-(((1)/(2)x-2)^6+5)} \, dx=
(64)/(3) or about 21.33333333
User Webber
by
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