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IgorLugansk [536]
3 years ago
13

Evaluating integral by net area

Mathematics
1 answer:
N76 [4]3 years ago
8 0

The graph of f(x)=2 is a horizontal line. So the area under f(x) on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.

The graph of g(x)=\sqrt{64-x^2} is the upper half of a circle with radius \sqrt{64}=8. It's symmetric about x=0, so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is \frac{\pi 8^2}4=16\pi.

Then use the fact that the definite integral is distributive over sums, meaning

\displaystyle\int_0^8f(x)+g(x)\,\mathrm dx=\int_0^8f(x)\,\mathrm dx+\int_0^8g(x)\,\mathrm dx=16+16\pi

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