What is the integral from 0 to 1 of 4/3 t^{3/2}(t+2) ?

The integral from 0 to 1 of 4/3 t^{3/2}(t+2) is 152/105

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integral from 0 to 1 of 4/3 t^{3/2}(t+2)

\int_{0}^{1} \frac{4}{3} t^{\frac{3}{2}} (t + 2) d t

\frac{152}{105}

Decimal Notation

1.44761 \dots

Solution steps

\int_{0}^{1} \frac{4}{3} t^{\frac{3}{2}} (t + 2) d t

Take the constant out:

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

= \frac{4}{3} \cdot \int_{0}^{1} t^{\frac{3}{2}} (t + 2) d t

Apply Integration By Parts

: [\frac{2}{5} t^{\frac{5}{2}} (t + 2) - \int \frac{2}{5} t^{\frac{5}{2}} d t]_{0}^{1}

= \frac{4}{3} [\frac{2}{5} t^{\frac{5}{2}} (t + 2) - \int \frac{2}{5} t^{\frac{5}{2}} d t]_{0}^{1}

\int \frac{2}{5} t^{\frac{5}{2}} d t = \frac{4}{35} t^{\frac{7}{2}}

= \frac{4}{3} [\frac{2}{5} t^{\frac{5}{2}} (t + 2) - \frac{4}{35} t^{\frac{7}{2}}]_{0}^{1}

Simplify

\frac{4}{3} [\frac{2}{5} t^{\frac{5}{2}} (t + 2) - \frac{4}{35} t^{\frac{7}{2}}]_{0}^{1} : \frac{4}{3} [\frac{10 t ^{\frac{7}{2}} + 28 t ^{\frac{5}{2}}}{35}]_{0}^{1}

= \frac{4}{3} [\frac{10 t ^{\frac{7}{2}} + 28 t ^{\frac{5}{2}}}{35}]_{0}^{1}

Compute the boundaries:

\frac{38}{35}

= \frac{4}{3} \cdot \frac{38}{35}

Simplify

= \frac{152}{105}

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What is the integral from 0 to 1 of 4/3 t^{3/2}(t+2) ?
The integral from 0 to 1 of 4/3 t^{3/2}(t+2) is 152/105