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

The integral from 0 to 1 of 6x^2sqrt(1-x^3) is 4/3

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{0} - \frac{u}{3} d u

= 6 \cdot \int_{1}^{0} - \frac{u}{3} d u

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

= 6 (- \int_{0}^{1} - \frac{u}{3} d u)

Take the constant out:

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

= 6 (- (- \frac{1}{3} \cdot \int_{0}^{1} u d u))

Apply the Power Rule

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

= 6 (- (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}))

Simplify

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

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

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{4}{3}

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What is the integral from 0 to 1 of 6x^2sqrt(1-x^3) ?
The integral from 0 to 1 of 6x^2sqrt(1-x^3) is 4/3