Find the exact value of the integral using formulas from geometry.\\ 6 \(\int_0^6 \sqrt{36 - x^2} dx\) 6 \(\int_0^6 \sqrt{36 - x^2} dx = \) (Type an exact answer, using \(\pi\) as needed.)
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In this case, the function is constant at 6 over the interval [6, 6], so the integral is simply the width of the interval times the height of the function. Show more…
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