What is the integral from-3 to 3 of (9-x^2)^{3/2} ?

The integral from-3 to 3 of (9-x^2)^{3/2} is (243pi)/8

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integral from-3 to 3 of (9-x^2)^{3/2}

\int_{- 3}^{3} (9 - x^{2})^{\frac{3}{2}} d x

\frac{243 π}{8}

Decimal

95.42587 \dots

Solution steps

\int_{- 3}^{3} (9 - x^{2})^{\frac{3}{2}} d x

Apply Integration By Parts

= [x (9 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} (9 - x^{2})^{\frac{1}{2}} d x]_{- 3}^{3}

\int - 3 x^{2} (9 - x^{2})^{\frac{1}{2}} d x = - 3 (\frac{1}{4} x^{3} (9 - x^{2})^{\frac{1}{2}} + \frac{81}{8} arcsin (\frac{1}{3} x) - \frac{81}{16} sin (2 arcsin (\frac{1}{3} x)))

= [x (9 - x^{2})^{\frac{3}{2}} - (- 3 (\frac{1}{4} x^{3} (9 - x^{2})^{\frac{1}{2}} + \frac{81}{8} arcsin (\frac{1}{3} x) - \frac{81}{16} sin (2 arcsin (\frac{1}{3} x))))]_{- 3}^{3}

Simplify

= [x (9 - x^{2})^{\frac{3}{2}} + 3 (\frac{1}{4} x^{3} (9 - x^{2})^{\frac{1}{2}} + \frac{81}{8} arcsin (\frac{1}{3} x) - \frac{81}{16} sin (2 arcsin (\frac{1}{3} x)))]_{- 3}^{3}

Compute the boundaries:

\frac{243 π}{8}

= \frac{243 π}{8}

What is the integral from-3 to 3 of (9-x^2)^{3/2} ?
The integral from-3 to 3 of (9-x^2)^{3/2} is (243pi)/8