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

The integral from 0 to 2 of t^3sqrt(3t^2+1) is (2(221sqrt(13)+1))/(135)

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

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

\frac{2 ( 221 13 + 1 )}{135}

Decimal Notation

11.81965 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{13} \frac{u ( u - 1 )}{18} d u

= \int_{1}^{13} \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}^{13} u (u - 1) d u

Expand

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

= \frac{1}{18} \cdot \int_{1}^{13} 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}^{13} u^{\frac{3}{2}} d u - \int_{1}^{13} u d u)

\int_{1}^{13} u^{\frac{3}{2}} d u = \frac{338 13 - 2}{5}

\int_{1}^{13} u d u = \frac{26 13 - 2}{3}

= \frac{1}{18} (\frac{338 13 - 2}{5} - \frac{26 13 - 2}{3})

Simplify

\frac{1}{18} (\frac{338 13 - 2}{5} - \frac{26 13 - 2}{3}) : \frac{2 ( 221 13 + 1 )}{135}

= \frac{2 ( 221 13 + 1 )}{135}

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What is the integral from 0 to 2 of t^3sqrt(3t^2+1) ?
The integral from 0 to 2 of t^3sqrt(3t^2+1) is (2(221sqrt(13)+1))/(135)