Evaluate the integral: ${\int }_1^7{w}^2\ln \left(w\right) dw$

• A. $[w^2\ln (w)-w^2]\mid 1,7$
• B. $\frac{1}{2}[w^2\ln (w)-\frac{w^2}{2}]\mid 1,7$
• C. ${\int }_1^7\ln \left(w\right) dw=[w\ln (w)-w]\mid 1,7$
• D. $\frac{1}{3}[w^3\ln (w)-\frac{w^3}{3}]\mid 1,7$

Answer: (D)

Verified step by step guidance

1. Step 1: Recognize that the integral \( \int_1^7 w^2 \ln(w) \, dw \) involves a product of functions \( w^2 \) and \( \ln(w) \). This suggests using integration by parts, which is given by \( \int u \, dv = uv - \int v \, du \).
2. Step 2: Choose \( u = \ln(w) \) (since its derivative simplifies) and \( dv = w^2 \, dw \). Compute \( du = \frac{1}{w} \, dw \) and \( v = \frac{w^3}{3} \) (by integrating \( w^2 \)).
3. Step 3: Substitute into the integration by parts formula: \( \int w^2 \ln(w) \, dw = \frac{w^3}{3} \ln(w) - \int \frac{w^3}{3} \cdot \frac{1}{w} \, dw \). Simplify the second integral to \( \int \frac{w^3}{3} \cdot \frac{1}{w} \, dw = \int \frac{w^2}{3} \, dw \).
4. Step 4: Compute \( \int \frac{w^2}{3} \, dw \). Factor out \( \frac{1}{3} \) and integrate \( w^2 \) to get \( \frac{1}{3} \cdot \frac{w^3}{3} = \frac{w^3}{9} \). Substitute this result back into the expression from Step 3.
5. Step 5: Combine terms to express the integral as \( \frac{w^3}{3} \ln(w) - \frac{w^3}{9} \). Evaluate this expression at the bounds \( w = 7 \) and \( w = 1 \) to find the definite integral value.