What is the integral from sqrt(y) to 1 of yln(x) ?

The integral from sqrt(y) to 1 of yln(x) is y(-1-(sqrt(y)ln(y))/2+sqrt(y))

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integral from sqrt(y) to 1 of yln(x)

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

y (- 1 - \frac{y ln ( y )}{2} + y)

Solution steps

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

Take the constant out:

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

= y \cdot \int_{y}^{1} ln (x) d x

Apply Integration By Parts

: [x ln (x) - \int 1 d x]_{y}^{1}

= y [x ln (x) - \int 1 d x]_{y}^{1}

\int 1 d x = x

= y [x ln (x) - x]_{y}^{1}

Compute the boundaries:

- 1 - \frac{1}{2} y ln (y) + y

= y (- 1 - \frac{1}{2} y ln (y) + y)

Simplify

= y (- 1 - \frac{y ln ( y )}{2} + y)

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What is the integral from sqrt(y) to 1 of yln(x) ?
The integral from sqrt(y) to 1 of yln(x) is y(-1-(sqrt(y)ln(y))/2+sqrt(y))