Edexcel IAL October 2021 P4 (WMA14/01) Q8 Integration by Parts, Volumes of Revolution
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Solutions relying on calculator technology are not acceptable.
(a) Find \(\int x^2 \ln x dx\)(3)
Figure 3 shows a sketch of part of the curve with equation
\(y = x \ln x \quad x > 0\)
The region \(R\), shown shaded in Figure 3, lies entirely above the x-axis and is bounded by the curve, the x-axis and the line with equation \(x = e\).
This region is rotated through \(2\pi radians\) about the x-axis to form a solid of revolution.
(b) Find the exact volume of the solid formed, giving your answer in simplest form.(4)
Solution to Part a: Integrate \(\int x^2 \ln x dx\)
Key Concepts Used:
- Integration by Parts: \(\int u dv = uv – \int v du\).
- Choice of \(u\) and \(dv\): Let \(u = \ln x\) (since its derivative simplifies) and \(dv = x^2 dx\).
Step-by-Step Solution:
Let \( I = \int x^2 \ln x dx \)
-
Apply integration by parts: Let
\( u = \ln x, \quad dv = x^2 dx \)
Then
\( du = \frac{1}{x} dx, \quad v = \frac{x^3}{3} \)So,
\( I = uv – \int vdu = \ln x \cdot \frac{x^3}{3} – \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x – \frac{1}{3} \int x^2 dx \) -
Now integrate \(\int x^2 dx\):
\( \int x^2 dx = \frac{x^3}{3} \)
Thus,
\( I = \frac{x^3}{3} \ln x – \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x – \frac{x^3}{9} + C \)
Final Answer:
\( \frac{x^3}{3} \ln x – \frac{x^3}{9} + C \)
Solution to Part b: Volume of Solid of Revolution
Key Concepts Used:
- Volume of Revolution: \( V = \pi \int_a^b y^2 dx \) for rotation about the x-axis.
- Integration by Parts: Will be applied twice due to the form of \( y^2 \).
- Limits of Integration: From \( x = 1 \) to \( x = e \) (since \( y = x \ln x = 0 \) at \( x = 1 \)).
Step-by-Step Solution:
-
Given curve:
\(y = x \ln x\)
So,
\(y^2 = (x \ln x)^2 = x^2 (\ln x)^2\) -
Volume integral:
\(V = \pi \int_1^e y^2 dx = \pi \int_1^e x^2 (\ln x)^2 dx\)
-
Let \( I_1 = \int x^2 (\ln x)^2 dx \):
Use integration by parts.
Let\(u = (\ln x)^2, \quad dv = x^2 dx\)Then
\(du = 2 \ln x \cdot \frac{1}{x} dx = \frac{2 \ln x}{x} dx, \quad v = \frac{x^3}{3}\)So,
\(I_1 = \frac{x^3}{3} (\ln x)^2 – \int \frac{x^3}{3} \cdot \frac{2 \ln x}{x} dx = \frac{x^3}{3} (\ln x)^2 – \frac{2}{3} \int x^2 \ln x dx\) -
From part (a), \( \int x^2 \ln x dx = \frac{x^3}{3} \ln x – \frac{x^3}{9} \):
So,\(I_1 = \frac{x^3}{3} (\ln x)^2 – \frac{2}{3} \left( \frac{x^3}{3} \ln x – \frac{x^3}{9} \right) + C\)\(= \frac{x^3}{3} (\ln x)^2 – \frac{2x^3}{9} \ln x + \frac{2x^3}{27} + C\) -
Now evaluate definite integral from 1 to e:
\(\int_{1}^{e} x^2 (\ln x)^2 dx = \left[ \frac{x^3}{3} (\ln x)^2 – \frac{2x^3}{9} \ln x + \frac{2x^3}{27} \right]_{1}^{e}\)
At \( x = e \):
\(\ln e = 1\)
\(\frac{e^3}{3} (1)^2 – \frac{2e^3}{9} (1) + \frac{2e^3}{27} = \frac{e^3}{3} – \frac{2e^3}{9} + \frac{2e^3}{27}\)Find common denominator (27):
\(= \frac{9e^3}{27} – \frac{6e^3}{27} + \frac{2e^3}{27} = \frac{5e^3}{27}\)At \( x = 1 \):
\(\ln 1 = 0\)
\(\frac{1}{3} (0) – \frac{2}{9} (0) + \frac{2}{27} = \frac{2}{27}\)Therefore,
\(\int_{1}^{e} x^2 (\ln x)^2 dx = \frac{5e^3}{27} – \frac{2}{27} = \frac{5e^3 – 2}{27}\) -
Volume:
\(V = \pi \cdot \frac{5e^3 – 2}{27} = \frac{\pi}{27} (5e^3 – 2)\)
Final Answer
\(\frac{\pi}{27} (5e^3 – 2)\)