What is the integral from 0 to 1 of x^2sqrt(6-2x^3) ?

The integral from 0 to 1 of x^2sqrt(6-2x^3) is (6sqrt(6)-8)/9

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integral from 0 to 1 of x^2sqrt(6-2x^3)

\int_{0}^{1} x^{2} 6 - 2 x^{3} d x

\frac{6 6 - 8}{9}

Decimal

0.74410 \dots

Solution steps

\int_{0}^{1} x^{2} 6 - 2 x^{3} d x

x^{2} 6 - 2 x^{3} = 2 x^{2} 3 - x^{3}

= \int_{0}^{1} 2 x^{2} 3 - x^{3} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x^{2} 3 - x^{3} d x

Apply u-substitution

: \int_{3}^{2} - \frac{u}{3} d u

= 2 \cdot \int_{3}^{2} - \frac{u}{3} d u

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

= 2 (- \int_{2}^{3} - \frac{u}{3} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{3} \cdot \int_{2}^{3} u d u))

Apply the Power Rule

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

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

Simplify

2 (- (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{3})) : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{3}

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

Compute the boundaries:

23 - \frac{4 2}{3}

= \frac{2}{3} (23 - \frac{4 2}{3})

Simplify

= \frac{6 6 - 8}{9}

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What is the integral from 0 to 1 of x^2sqrt(6-2x^3) ?
The integral from 0 to 1 of x^2sqrt(6-2x^3) is (6sqrt(6)-8)/9