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

The integral from 0 to 1 of t^2sqrt(t^3+3) is (2(8-3sqrt(3)))/9

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

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

\frac{2 ( 8 - 3 3 )}{9}

Decimal

0.62307 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{8}{3} - 3

= \frac{2}{3} (\frac{8}{3} - 3)

Simplify

= \frac{2 ( 8 - 3 3 )}{9}

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What is the integral from 0 to 1 of t^2sqrt(t^3+3) ?
The integral from 0 to 1 of t^2sqrt(t^3+3) is (2(8-3sqrt(3)))/9