What is the integral from 1 to 2 of 9r^4ln(r) ?

The integral from 1 to 2 of 9r^4ln(r) is 9(-31/25+32/5 ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of 9r^4ln(r)

\int_{1}^{2} 9 r^{4} ln (r) d r

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

Decimal

28.76527 \dots

Solution steps

\int_{1}^{2} 9 r^{4} ln (r) d r

Take the constant out:

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

= 9 \cdot \int_{1}^{2} r^{4} ln (r) d r

Apply Integration By Parts

= 9 [\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}

= 9 [\frac{1}{5} r^{5} ln (r) - \frac{r ^{5}}{25}]_{1}^{2}

Compute the boundaries:

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

= 9 (- \frac{31}{25} + \frac{32}{5} ln (2))

What is the integral from 1 to 2 of 9r^4ln(r) ?
The integral from 1 to 2 of 9r^4ln(r) is 9(-31/25+32/5 ln(2))