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

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

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

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

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{2}{9}

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