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

The integral from e to 1 of sqrt(t)ln(t) is -(2e^{3/2}+4)/9

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

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

- \frac{2 e ^{\frac{3}{2}} + 4}{9}

Decimal Notation

- 1.44037 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{e} t ln (t) d t

Apply Integration By Parts

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

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

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

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

3 \cdot \frac{4}{9} t^{\frac{3}{2}} = \frac{4}{3} t^{\frac{3}{2}}

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

Compute the boundaries:

\frac{2 e ^{\frac{3}{2}}}{9} + \frac{4}{9}

= - (\frac{2 e ^{\frac{3}{2}}}{9} + \frac{4}{9})

Simplify

= - \frac{2 e ^{\frac{3}{2}} + 4}{9}

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