Question

Evaluate the definite integral. $$ \int_{1}^{3} x^{2} \ln x d x $$

Short answer

The value of the definite integral \( \int_{1}^{3} x^{2} \ln x dx \) is given by \( [\frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3}]_{1}^{3} \), which when calculated will yield the final numerical answer.

Step-by-step solution

Using the formula for integration by parts.

We choose \( u = x^{2} \) and \( dv = \ln x dx \). Hence, \( du = 2x dx \) and \( v = \int \ln x dx = \int 1 \cdot \ln x dx = x \ln x - x \). The formula for integration by parts gives a new integral: \( \int u dv = uv - \int v du = [x^{2} (x \ln x - x)] - \int (x \ln x - x) 2x dx = x^{3} \ln x - x^{3} - 2 \int x^{2} \ln x dx + 2 \int x^{2} dx \).

Solving for the integral.

The integral \( \int x^{2} \ln x dx \) appears on both sides of the equation. Therefore, let's add \( 2 \int x^{2} \ln x dx \) to both sides to get \( 3 \int x^{2} \ln x dx = x^{3} \ln x - x^{3} + 2 \int x^{2} dx \). Hence, \( \int x^{2} \ln x dx = \frac{1}{3} (x^{3} \ln x - x^{3} + 2 \int x^{2} dx) \). The integration of \( x^{2} \) is straightforward. We get \( \int x^{2} dx = \frac{1}{3} x^{3} \). Substituting this into the equation gives \( \int x^{2} \ln x dx = \frac{1}{3} (x^{3} \ln x - x^{3} + 2(\frac{1}{3} x^{3})) \). This simplifies to \( \int x^{2} \ln x dx = \frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3} \).

Evaluating the definite integral.

The definite integral of \( x^{2} \ln x \) from \( x = 1 \) to \( x = 3 \) is given by \( \int_{1}^{3} x^{2} \ln x dx = [\frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3}]_{1}^{3} \). After evaluating this expression at \( x = 3 \) and at \( x = 1 \) and subtracting, you will obtain the final answer.