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

The integral from 1 to 4 of yln(y) is 16ln(2)-15/4

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integral from 1 to 4 of yln(y)

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

16 ln (2) - \frac{15}{4}

Decimal Notation

7.34035 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{y}{2} d y = \frac{y ^{2}}{4}

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

Compute the boundaries:

16 ln (2) - \frac{15}{4}

= 16 ln (2) - \frac{15}{4}

n = 0 \sum \infty (\frac{e}{2})^{n}

\int x ln (x + 3) d x

\int_{1}^{e} 8 t^{2} ln (t) d t

f (x) = x^{2} + 6 x - 1

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 5

What is the integral from 1 to 4 of yln(y) ?
The integral from 1 to 4 of yln(y) is 16ln(2)-15/4