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

The integral from 0 to 1 of x^3sqrt(1+3x^2) is 58/135

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

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

\frac{58}{135}

Decimal Notation

0.42962 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{u ( u - 1 )}{18} d u

= \int_{1}^{4} \frac{u ( u - 1 )}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int_{1}^{4} u (u - 1) d u

Expand

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

= \frac{1}{18} \cdot \int_{1}^{4} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{18} (\int_{1}^{4} u^{\frac{3}{2}} d u - \int_{1}^{4} u d u)

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= \frac{1}{18} (\frac{62}{5} - \frac{14}{3})

Simplify

\frac{1}{18} (\frac{62}{5} - \frac{14}{3}) : \frac{58}{135}

= \frac{58}{135}

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What is the integral from 0 to 1 of x^3sqrt(1+3x^2) ?
The integral from 0 to 1 of x^3sqrt(1+3x^2) is 58/135