What is the integral from 1 to 5 of 7r^2ln(r) ?

The integral from 1 to 5 of 7r^2ln(r) is 7(-124/9+125/3 ln(5))

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integral from 1 to 5 of 7r^2ln(r)

\int_{1}^{5} 7 r^{2} ln (r) d r

7 (- \frac{124}{9} + \frac{125}{3} ln (5))

Solution steps

\int_{1}^{5} 7 r^{2} ln (r) d r

Take the constant out:

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

= 7 \cdot \int_{1}^{5} 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}^{5}

= 7 [\frac{1}{3} r^{3} ln (r) - \int \frac{r ^{2}}{3} d r]_{1}^{5}

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

= 7 [\frac{1}{3} r^{3} ln (r) - \frac{r ^{3}}{9}]_{1}^{5}

Compute the boundaries:

- \frac{124}{9} + \frac{125}{3} ln (5)

= 7 (- \frac{124}{9} + \frac{125}{3} ln (5))

x \to - 2 lim (\frac{3}{x ^{2} - 4})

\int \frac{x - 2}{x ^{3}} d x

\frac{d}{d x} ((\frac{- 2 x - 1}{- 4 x + 2})^{5})

\int_{2}^{4} 2 x^{2} - 3 x + 1 d x

\frac{\partial}{\partial x} (cos (x ln (y)))

What is the integral from 1 to 5 of 7r^2ln(r) ?
The integral from 1 to 5 of 7r^2ln(r) is 7(-124/9+125/3 ln(5))