What is the integral from 1 to 3 of 11r^3ln(r) ?

The integral from 1 to 3 of 11r^3ln(r) is 11(-5+81/4 ln(3))

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integral from 1 to 3 of 11r^3ln(r)

\int_{1}^{3} 11 r^{3} ln (r) d r

11 (- 5 + \frac{81}{4} ln (3))

Decimal

189.71588 \dots

Solution steps

\int_{1}^{3} 11 r^{3} ln (r) d r

Take the constant out:

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

= 11 \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}

= 11 [\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}

= 11 [\frac{1}{4} r^{4} ln (r) - \frac{r ^{4}}{16}]_{1}^{3}

Compute the boundaries:

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

= 11 (- 5 + \frac{81}{4} ln (3))

y^{^{''}} + 25 y = 40 sin (5 t)

\frac{\partial}{\partial y} (x y + y)

10 e^{(- 10 t)}

\int e^{4 x} x d x

f (x) = 4 6 x^{2} + 7

What is the integral from 1 to 3 of 11r^3ln(r) ?
The integral from 1 to 3 of 11r^3ln(r) is 11(-5+81/4 ln(3))