What is the integral from 1 to 3 of 5r^3ln(r) ?

The integral from 1 to 3 of 5r^3ln(r) is 5(-5+81/4 ln(3))

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integral from 1 to 3 of 5r^3ln(r)

\int_{1}^{3} 5 r^{3} ln (r) d r

5 (- 5 + \frac{81}{4} ln (3))

Decimal Notation

86.23449 \dots

Solution steps

\int_{1}^{3} 5 r^{3} 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}^{3} r^{3} ln (r) d r

Apply Integration By Parts

: [\frac{1}{4} r^{4} ln (r) - \int \frac{r ^{3}}{4} d r]_{1}^{3}

= 5 [\frac{1}{4} r^{4} ln (r) - \int \frac{r ^{3}}{4} d r]_{1}^{3}

\int \frac{r ^{3}}{4} d r = \frac{r ^{4}}{16}

= 5 [\frac{1}{4} r^{4} ln (r) - \frac{r ^{4}}{16}]_{1}^{3}

Compute the boundaries:

- 5 + \frac{81}{4} ln (3)

= 5 (- 5 + \frac{81}{4} ln (3))

y^{^{''}} - y^{^{'}} - 2 y = 2 e^{- t}

\int tan^{3} (x) sec^{\frac{1}{2}} (x) d x

x \to 0 lim (\frac{1}{e ^{x}})

x \to 6 lim (\frac{x}{x ^{2} - 36})

\frac{d x}{d t} + x = - e^{t}

What is the integral from 1 to 3 of 5r^3ln(r) ?
The integral from 1 to 3 of 5r^3ln(r) is 5(-5+81/4 ln(3))