What is the integral from 1 to 4 of sqrt(y)ln(y) ?

The integral from 1 to 4 of sqrt(y)ln(y) is -28/9+32/3 ln(2)

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integral from 1 to 4 of sqrt(y)ln(y)

\int_{1}^{4} y ln (y) d y

- \frac{28}{9} + \frac{32}{3} ln (2)

Decimal Notation

4.28245 \dots

Solution steps

\int_{1}^{4} y ln (y) d y

Apply Integration By Parts

: [\frac{2}{3} y^{\frac{3}{2}} ln (y) - \int \frac{2}{3} y^{\frac{1}{2}} d y]_{1}^{4}

= [\frac{2}{3} y^{\frac{3}{2}} ln (y) - \int \frac{2}{3} y^{\frac{1}{2}} d y]_{1}^{4}

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

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

Compute the boundaries:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - \frac{28}{9} + \frac{32}{3} ln (2)

\frac{d x}{d t} = k x (1000 - x)

y^{^{''}} + 400 y = sec (20 x), y (0) = 5, y^{^{'}} (0) = - 5

\int (x - 1) sec^{2} (x) d x

(x - 1) ln (x)

\frac{\partial}{\partial x} (\frac{1}{3 x + 2})

What is the integral from 1 to 4 of sqrt(y)ln(y) ?
The integral from 1 to 4 of sqrt(y)ln(y) is -28/9+32/3 ln(2)