What is the integral from 1 to 2 of 2r^4ln(r) ?

The integral from 1 to 2 of 2r^4ln(r) is 2(-31/25+32/5 ln(2))

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integral from 1 to 2 of 2r^4ln(r)

\int_{1}^{2} 2 r^{4} ln (r) d r

2 (- \frac{31}{25} + \frac{32}{5} ln (2))

Decimal Notation

6.39228 \dots

Solution steps

\int_{1}^{2} 2 r^{4} ln (r) d r

Take the constant out:

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

= 2 \cdot \int_{1}^{2} r^{4} ln (r) d r

Apply Integration By Parts

: [\frac{1}{5} r^{5} ln (r) - \int \frac{r ^{4}}{5} d r]_{1}^{2}

= 2 [\frac{1}{5} r^{5} ln (r) - \int \frac{r ^{4}}{5} d r]_{1}^{2}

\int \frac{r ^{4}}{5} d r = \frac{r ^{5}}{25}

= 2 [\frac{1}{5} r^{5} ln (r) - \frac{r ^{5}}{25}]_{1}^{2}

Compute the boundaries:

- \frac{31}{25} + \frac{32}{5} ln (2)

= 2 (- \frac{31}{25} + \frac{32}{5} ln (2))

x \to 2 lim (csc (\frac{π x}{6}))

\frac{d ^{10}}{d x ^{10}} ((2 x + 5)^{7})

\int \frac{x ^{2}}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (e^{- x} - e^{- 2 x})

x \to 0 lim ([x])

What is the integral from 1 to 2 of 2r^4ln(r) ?
The integral from 1 to 2 of 2r^4ln(r) is 2(-31/25+32/5 ln(2))