What is the integral from 1 to 3 of 17r^3ln(r) ?

The integral from 1 to 3 of 17r^3ln(r) is 17(-5+81/4 ln(3))

Popular >

integral from 1 to 3 of 17r^3ln(r)

\int_{1}^{3} 17 r^{3} ln (r) d r

17 (- 5 + \frac{81}{4} ln (3))

Decimal Notation

293.19728 \dots

Solution steps

\int_{1}^{3} 17 r^{3} ln (r) d r

Take the constant out:

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

= 17 \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}

= 17 [\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}

= 17 [\frac{1}{4} r^{4} ln (r) - \frac{r ^{4}}{16}]_{1}^{3}

Compute the boundaries:

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

= 17 (- 5 + \frac{81}{4} ln (3))

y^{^{'}} - \frac{6 y}{x} = x^{6} e^{x}

\frac{d}{d y} (x y sin (y) + x y^{2} sin (x) + \frac{1}{x})

x \to 3 lim (csc (\frac{π x}{4}))

\int e^{- x} \cdot cos (x) d x

(x y^{3} + 1) d x + x^{2} y^{2} d y = 0

What is the integral from 1 to 3 of 17r^3ln(r) ?
The integral from 1 to 3 of 17r^3ln(r) is 17(-5+81/4 ln(3))