What is the integral from 0 to 2 of sqrt((4-x^2)^3) ?

The integral from 0 to 2 of sqrt((4-x^2)^3) is 3pi

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integral from 0 to 2 of sqrt((4-x^2)^3)

\int_{0}^{2} (4 - x^{2})^{3} d x

3 π

Decimal

9.42477 \dots

Solution steps

\int_{0}^{2} (4 - x^{2})^{3} d x

(4 - x^{2})^{3} = (4 - x^{2})^{\frac{3}{2}},

assuming

(4 - x^{2}) \geq 0

= \int_{0}^{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} d x]_{0}^{2}

= [x (4 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 4 - x^{2} d x]_{0}^{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})^{\frac{3}{2}} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))))]_{0}^{2}

Simplify

= [x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))]_{0}^{2}

Compute the boundaries:

3 π

= 3 π

\int_{- 2}^{0} \frac{1}{1 - 4 x} d x

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\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 2} d x

\int_{0}^{1} 2 x 1 - x^{2} d x

\int_{\frac{π}{4}}^{\frac{π}{2}} cot^{2} (θ) d θ

What is the integral from 0 to 2 of sqrt((4-x^2)^3) ?
The integral from 0 to 2 of sqrt((4-x^2)^3) is 3pi