QuestionCalculate the integral from 1 to 2 of $w^{2} \ln w \, dw$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the definite integral $\int_{1}^{} w^{} \ln w d w$ . We know the properties of logarithms and the rules of integration

STEP 2

We can solve this integral by using integration by parts, which is given by the formula $\int u dv = uv - \int v du$

STEP 3

We need to choose our $u$ and $dv$ for the integration by parts formula. A good rule of thumb is to choose $u$ to be a function that becomes simpler when we differentiate it. In this case, we choose $\ln w$ as $u$ and $w^2$ as $dv$ .
$u = \ln w, dv = w^2 dw$

STEP 4

Now, we need to find $du$ and $v$ . We differentiate $u$ to get $du$ and integrate $dv$ to get $v$ .
$du = \frac{1}{w} dw, v = \frac{w^3}{3}$

STEP 5

Now, we substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula.
$\int_{1}^{2} w^{2} \ln w d w = \left[\frac{w^3}{3} \ln w\right]_{1}^{2} - \int_{1}^{2} \frac{w^3}{3} \cdot \frac{1}{w} dw$

STEP 6

implify the integral on the right.
$\int_{1}^{2} w^{2} \ln w d w = \left[\frac{w^3}{3} \ln w\right]_{1}^{2} - \int_{1}^{2} \frac{w^2}{3} dw$

STEP 7

Now, we can evaluate the remaining integral.
$\int_{1}^{2} w^{2} \ln w d w = \left[\frac{w^3}{3} \ln w\right]_{1}^{2} - \left[\frac{w^3}{9}\right]_{1}^{2}$

STEP 8

Evaluate the definite integrals.
$\int_{1}^{2} w^{2} \ln w d w = \left[\frac{2^3}{3} \ln2 - \frac{1^3}{3} \ln1\right] - \left[\frac{2^3}{} - \frac{1^3}{}\right]$

STEP 9

implify the expression.
$\int_{}^{2} w^{2} \ln w d w = \left[\frac{8}{3} \ln2 -\right] - \left[\frac{8}{9} - \frac{}{9}\right]$

STEP 10

Further simplify the expression.
$\int_{}^{2} w^{2} \ln w d w = \frac{8}{3} \ln2 - \frac{7}{9}$ So, the value of the integral $\int_{}^{2} w^{2} \ln w d w$ is $\frac{8}{3} \ln2 - \frac{7}{9}$ .

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