What is the integral from 0 to 1 of x^2sqrt(4+x^3) ?

The integral from 0 to 1 of x^2sqrt(4+x^3) is (10sqrt(5)-16)/9

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integral from 0 to 1 of x^2sqrt(4+x^3)

\int_{0}^{1} x^{2} 4 + x^{3} d x

\frac{10 5 - 16}{9}

Decimal

0.70674 \dots

Solution steps

\int_{0}^{1} x^{2} 4 + x^{3} d x

Apply u-substitution

: \int_{4}^{5} \frac{u}{3} d u

= \int_{4}^{5} \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_{4}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{10 5}{3} - \frac{16}{3})

Simplify

= \frac{10 5 - 16}{9}

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What is the integral from 0 to 1 of x^2sqrt(4+x^3) ?
The integral from 0 to 1 of x^2sqrt(4+x^3) is (10sqrt(5)-16)/9