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Inga [223]
3 years ago
9

Find the range of the function f of x equals the integral from negative 6 to x of the square root of the quantity 36 minus t squ

ared
Mathematics
1 answer:
shtirl [24]3 years ago
7 0

f(x)=\displaystyle\int_{-6}^x\sqrt{36-t^2}\,\mathrm dt

The integrand is defined for 36-t^2\ge0, or -6\le t\le6, so the domain should be the same, -6\le x\le6.

When x=-6, the integral is 0.

The integrand is non-negative for all x in the domain, which means the value of f(x) increases monotonically over this domain. When x=6, the integral gives the area of the semicircle centered at the origin with radius 6, which is \dfrac\pi26^2=18\pi, so the range is \boxed{0\le f(x)\le 18\pi}.

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