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Evaluate the integral ∫(2 to 1) w² ln(w) dw?

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Final answer:

To evaluate the given integral, we can use integration by parts. After applying the integration by parts formula, we find that the integral is equal to (1/3)w³ ln(w) - (1/9)w³ + C. Substituting the upper and lower limits gives the final result: (1/3) - (8/9) ln(2) + (8/9).

Step-by-step explanation:

To evaluate the integral ∫(2 to 1) w² ln(w) dw, we can use the technique of integration by parts. Let's choose u = ln(w) and dv = w² dw. This gives du = 1/w dw and v = (1/3)w³. Applying the integration by parts formula, we have:

∫(2 to 1) w² ln(w) dw = uv - ∫v du

∫(2 to 1) w² ln(w) dw = [(1/3)w³ ln(w)] - ∫[(1/3)w³ (1/w) dw]

∫(2 to 1) w² ln(w) dw = (1/3)w³ ln(w) - (1/3)∫w² dw

∫(2 to 1) w² ln(w) dw = (1/3)w³ ln(w) - (1/3) * [(1/3)w³] + C

∫(2 to 1) w² ln(w) dw = (1/3)w³ ln(w) - (1/9)w³ + C

To find the definite integral, we substitute the upper and lower limits:

∫(2 to 1) w² ln(w) dw = [(1/3)(1)³ ln(1) - (1/9)(1)³] - [(1/3)(2)³ ln(2) - (1/9)(2)³]

∫(2 to 1) w² ln(w) dw = (1/3) - (8/9) ln(2) + (8/9)

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