What is the integral from 1 to 2 of sqrt(t)ln(t) ?

The integral from 1 to 2 of sqrt(t)ln(t) is (4sqrt(2))/3 ln(2)+(-8sqrt(2)+4)/9

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integral from 1 to 2 of sqrt(t)ln(t)

\int_{1}^{2} t ln (t) d t

\frac{4 2}{3} ln (2) + \frac{- 8 2 + 4}{9}

Decimal

0.49437 \dots

Solution steps

\int_{1}^{2} t ln (t) d t

Apply Integration By Parts

: [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{2}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{2}

\int \frac{2}{3} t^{\frac{1}{2}} d t = \frac{4}{9} t^{\frac{3}{2}}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \frac{4}{9} t^{\frac{3}{2}}]_{1}^{2}

Compute the boundaries:

\frac{4 2}{3} ln (2) - \frac{8 2}{9} + \frac{4}{9}

= \frac{4 2}{3} ln (2) - \frac{8 2}{9} + \frac{4}{9}

Simplify

= \frac{4 2}{3} ln (2) + \frac{- 8 2 + 4}{9}

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What is the integral from 1 to 2 of sqrt(t)ln(t) ?
The integral from 1 to 2 of sqrt(t)ln(t) is (4sqrt(2))/3 ln(2)+(-8sqrt(2)+4)/9