What is the integral from-4 to 4 of x^2sqrt(64-x^3) ?

The integral from-4 to 4 of x^2sqrt(64-x^3) is (2048sqrt(2))/9

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integral from-4 to 4 of x^2sqrt(64-x^3)

\int_{- 4}^{4} x^{2} 64 - x^{3} d x

\frac{2048 2}{9}

Solution steps

\int_{- 4}^{4} x^{2} 64 - x^{3} d x

Apply u-substitution

: \int_{128}^{0} - \frac{u}{3} d u

= \int_{128}^{0} - \frac{u}{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{128} - \frac{u}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{3} \cdot \int_{0}^{128} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{128}

= - (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{128})

Simplify

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{128}

Compute the boundaries:

\frac{2048 2}{3}

= \frac{1}{3} \cdot \frac{2048 2}{3}

Simplify

= \frac{2048 2}{9}

\int_{0}^{0.5} e^{x - 1} d x

\int_{0}^{0.5} 2 d x

\int_{0}^{10} e^{- x} d x

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

\int_{1}^{3} e^{x} - \frac{1}{x} d x

What is the integral from-4 to 4 of x^2sqrt(64-x^3) ?
The integral from-4 to 4 of x^2sqrt(64-x^3) is (2048sqrt(2))/9