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

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

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integral from-2 to 2 of (sqrt(4-x^2))^3

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

6 π

Decimal Notation

18.84955 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - 3 x^{2} 4 - x^{2} d x = - 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= [x (4 - x^{2})^{3} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))))]_{- 2}^{2}

Simplify

[x (4 - x^{2})^{3} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))))]_{- 2}^{2} : [x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))]_{- 2}^{2}

= [x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 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 (sqrt(4-x^2))^3 ?
The integral from-2 to 2 of (sqrt(4-x^2))^3 is 6pi