What is the integral from 0 to 2 of sqrt(t)(2t-2) ?

The integral from 0 to 2 of sqrt(t)(2t-2) is (8sqrt(2))/(15)

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integral from 0 to 2 of sqrt(t)(2t-2)

\int_{0}^{2} t (2 t - 2) d t

\frac{8 2}{15}

Solution steps

\int_{0}^{2} t (2 t - 2) d t

Apply Integration By Parts

: [\frac{2}{3} t^{\frac{3}{2}} (2 t - 2) - \int \frac{4}{3} t^{\frac{3}{2}} d t]_{0}^{2}

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

\int \frac{4}{3} t^{\frac{3}{2}} d t = \frac{8}{15} t^{\frac{5}{2}}

= [\frac{2}{3} t^{\frac{3}{2}} (2 t - 2) - \frac{8}{15} t^{\frac{5}{2}}]_{0}^{2}

Simplify

[\frac{2}{3} t^{\frac{3}{2}} (2 t - 2) - \frac{8}{15} t^{\frac{5}{2}}]_{0}^{2} : [\frac{1}{15} (12 t^{\frac{5}{2}} - 20 t^{\frac{3}{2}})]_{0}^{2}

= [\frac{1}{15} (12 t^{\frac{5}{2}} - 20 t^{\frac{3}{2}})]_{0}^{2}

Compute the boundaries:

\frac{8 2}{15}

= \frac{8 2}{15}

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What is the integral from 0 to 2 of sqrt(t)(2t-2) ?
The integral from 0 to 2 of sqrt(t)(2t-2) is (8sqrt(2))/(15)