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

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

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

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

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

Decimal

12.78456 \dots

Solution steps

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

Take the constant out:

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

= 4 \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}

= 4 [\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}

= 4 [\frac{1}{5} r^{5} ln (r) - \frac{r ^{5}}{25}]_{1}^{2}

Compute the boundaries:

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

= 4 (- \frac{31}{25} + \frac{32}{5} ln (2))

y = 2 x + 1

y^{^{'}} + \frac{3}{x} y = 0, y (1) = 5

f (x) = e^{x} 1 + x^{2}

\frac{d}{d x} (ln (x^{2} + 2 - y))

\frac{\partial}{\partial x} (\frac{x}{sin ( 2 x ) + 3 y + z ^{2}})

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