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

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

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

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

\frac{2 ( 4 2 - 2 )}{9}

Decimal Notation

0.81263 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{u}{3} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{4 2}{3} - \frac{2}{3}

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

Simplify

= \frac{2 ( 4 2 - 2 )}{9}

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