What is the integral from-2 to 2 of (4-x^2)^{3/2} ?

The integral from-2 to 2 of (4-x^2)^{3/2} is 6pi

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integral from-2 to 2 of (4-x^2)^{3/2}

\int_{- 2}^{2} (4 - x^{2})^{\frac{3}{2}} d x

6 π

Decimal

18.84955 \dots

Solution steps

\int_{- 2}^{2} (4 - x^{2})^{\frac{3}{2}} d x

Apply Integration By Parts

: [x (4 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} (4 - x^{2})^{\frac{1}{2}} d x]_{- 2}^{2}

= [x (4 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} (4 - x^{2})^{\frac{1}{2}} d x]_{- 2}^{2}

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

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

Simplify

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

Compute the boundaries:

6 π

= 6 π

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What is the integral from-2 to 2 of (4-x^2)^{3/2} ?
The integral from-2 to 2 of (4-x^2)^{3/2} is 6pi