What is the integral from 1 to 2 of 8r^2ln(r) ?

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

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integral from 1 to 2 of 8r^2ln(r)

\int_{1}^{2} 8 r^{2} ln (r) d r

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

Decimal Notation

8.56491 \dots

Solution steps

\int_{1}^{2} 8 r^{2} ln (r) d r

Take the constant out:

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

= 8 \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}

= 8 [\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}

= 8 [\frac{1}{3} r^{3} ln (r) - \frac{r ^{3}}{9}]_{1}^{2}

Compute the boundaries:

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

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

\int 3 t + 2 d t

f (x) = \frac{x ^{3}}{3} - \frac{3}{x ^{3}}, at x = - 1

f (x) = (\frac{2 x}{7} + 1)

\int x^{2} sin^{2} (x) d x

x \to 2 - lim (x + 1)

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