A+b=3b 6a−3b=5b 2a=9b a−2b/a+2b = 12 2a−3b/5 = a+b/3
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The graph of is a horizontal line. So the area under
on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.
The graph of is the upper half of a circle with radius
. It's symmetric about
, so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is
.
Then use the fact that the definite integral is distributive over sums, meaning
