What is the integral from 1 to 2 of 5r^2ln(r) ?

The integral from 1 to 2 of 5r^2ln(r) is 5(-7/9+8/3 ln(2))

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integral from 1 to 2 of 5r^2ln(r)

\int_{1}^{2} 5 r^{2} ln (r) d r

5 (- \frac{7}{9} + \frac{8}{3} ln (2))

Decimal Notation

5.35307 \dots

Solution steps

\int_{1}^{2} 5 r^{2} ln (r) d r

Take the constant out:

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

= 5 \cdot \int_{1}^{2} r^{2} ln (r) d r

Apply Integration By Parts

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

= 5 [\frac{1}{3} r^{3} ln (r) - \int \frac{r ^{2}}{3} d r]_{1}^{2}

\int \frac{r ^{2}}{3} d r = \frac{r ^{3}}{9}

= 5 [\frac{1}{3} r^{3} ln (r) - \frac{r ^{3}}{9}]_{1}^{2}

Compute the boundaries:

- \frac{7}{9} + \frac{8}{3} ln (2)

= 5 (- \frac{7}{9} + \frac{8}{3} ln (2))

\frac{d}{d x} ((- 3 cos (2 x) - sin (3 x))^{2})

\frac{4 x}{9 - cot ( x )}

t \to 0 lim (e^{- 4 t})

\int \frac{( ln ( x ) ) ^{17}}{x} d x

\frac{d y}{d x} = \frac{4 y ^{3}}{x ^{2}}

What is the integral from 1 to 2 of 5r^2ln(r) ?
The integral from 1 to 2 of 5r^2ln(r) is 5(-7/9+8/3 ln(2))