A-Level Definite integrals: worked solution

Question

Region R is bounded by y=x ln x, the lines x=e and x=e², and the x-axis. When R is rotated through 2π about the x-axis, give the exact volume in the form (πe³/27)(ae³+b) for integers a and b.

Rotating a region about the x-axis gives the volume V = π ∫ y² dx between the two x-values. Here y = x ln x and x runs from e to e², so V = π ∫ from e to e² of (x ln x)² dx = π ∫ from e to e² of x²(ln x)² dx Work out the indefinite integral first, by parts. Choose the LOGARITHM as the part to differentiate: differentiating (ln x)² simplifies it, whereas integrating it does not. Take u = (ln x)² and dv/dx = x², so du/dx = 2(ln x)(1/x) and v = x³/3. ∫ x²(ln x)² dx = (x³/3)(ln x)² - ∫ (x³/3) · 2(ln x)/x dx = (x³/3)(ln x)² - (2/3) ∫ x² ln x dx The remaining integral still contains a logarithm, so apply parts a second time, with u = ln x and dv/dx = x², giving du/dx = 1/x and v = x³/3: ∫ x² ln x dx = (x³/3) ln x - ∫ (x³/3)(1/x) dx = (x³/3) ln x - ∫ x²/3 dx = (x³/3) ln x - x³/9 Substitute this back: ∫ x²(ln x)² dx = (x³/3)(ln x)² - (2/3)[ (x³/3) ln x - x³/9 ] = (x³/3)(ln x)² - (2x³/9) ln x + 2x³/27 Now apply the limits. At x = e²: ln x = 2 and x³ = (e²)³ = e⁶ (e⁶/3)(2²) - (2e⁶/9)(2) + 2e⁶/27 = (4/3)e⁶ - (4/9)e⁶ + (2/27)e⁶ Over the common denominator 27: (36e⁶ - 12e⁶ + 2e⁶)/27 = 26e⁶/27 At x = e: ln x = 1 and x³ = e³ (e³/3)(1²) - (2e³/9)(1) + 2e³/27 = (9e³ - 6e³ + 2e³)/27 = 5e³/27 Subtract the lower value from the upper: 26e⁶/27 - 5e³/27 = (1/27)(26e⁶ - 5e³) Take out the common factor e³ to reach the requested form: = (e³/27)(26e³ - 5) So V = (πe³/27)(26e³ - 5), giving a = 26 and b = -5.