Edexcel IAL June 2020 P4 (WMA14/01) Q5 Integration by Parts
(a) Find \(\int \frac{\ln x}{x^2} dx\)(4)
Figure 3 shows a sketch of part of the curve with equation
\( y = \frac{3 + 2x + \ln x}{x^2} \quad x > 0.5 \)
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 4\).
(b) Use the answer to part (a) to find the exact area of \(R\), writing your answer in simplest form.(4)
Solution to Part a: Integration by Parts
Key Concepts Used:
- Integration by Parts Formula: \(\int u dv = uv – \int v du\)
- Choice of \(u\) and \(dv\): Select \(u = \ln x\) to simplify upon differentiation.
- Power Rule Integration: \(\int x^{-2} dx = -x^{-1} + C\)
Step-by-Step Solution:
We want to find: \( I = \int \frac{\ln x}{x^2} dx = \int x^{-2} \ln x dx \)
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Apply Integration by Parts: Let:
\( u = \ln x \rightarrow du = \frac{1}{x} dx \)
\( dv = x^{-2} dx \rightarrow v = -x^{-1} = -\frac{1}{x} \)Then:
\( I = uv – \int v du = (\ln x) \left( -\frac{1}{x} \right) – \int \left( -\frac{1}{x} \right) \left( \frac{1}{x} \right) dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} dx \) -
Integrate \(\int \frac{1}{x^2} dx\):
\( \int x^{-2} dx = -x^{-1} = -\frac{1}{x} \)
Substitute Back:
\( I = -\frac{\ln x}{x} – \frac{1}{x} + C = -\frac{1}{x} (\ln x + 1) + C \)
Final Answer
\( -\frac{\ln x}{x} – \frac{1}{x} + C \)
Solution to Part b: Exact Area of R
Key Concepts Used:
- Definite Integral for Area: The area under \( y = \frac{3+2x+\ln x}{x^2} \) from \( x = 2 \) to \( x = 4 \).
- Splitting the Integral: Break into simpler parts: \(\int \frac{3}{x^2} dx, \int \frac{2x}{x^2} dx, \) and \(\int \frac{\ln x}{x^2} dx\).
- Using Part (a): Use the result from part (a) for \(\int \frac{\ln x}{x^2} dx\).
- Evaluation at Bounds: Substitute \( x = 2 \) and \( x = 4 \).
Step-by-Step Solution:
The area \( A \) is: \( A = \int_{2}^{4} \frac{3 + 2x + \ln x}{x^2} dx = \int_{2}^{4} (3x^{-2} + 2x^{-1} + x^{-2}\ln x) dx \)
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Split the integral:
\( A = \int_{2}^{4} 3x^{-2} dx + \int_{2}^{4} 2x^{-1} dx + \int_{2}^{4} x^{-2} \ln x dx \)
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Integrate each part:
1. \( \int 3x^{-2} dx = 3 \int x^{-2} dx = 3(-x^{-1}) = -\frac{3}{x} \)
2. \( \int 2x^{-1} dx = 2 \int \frac{1}{x} dx = 2\ln|x| \)
3. From part (a): \( \int x^{-2} \ln x dx = -\frac{\ln x}{x} – \frac{1}{x} \)So, the antiderivative is:
\( F(x) = -\frac{3}{x} + 2\ln x – \frac{\ln x}{x} – \frac{1}{x} = -\frac{4}{x} + 2\ln x – \frac{\ln x}{x} \) -
Evaluate from \( x = 2 \) to \( x = 4 \):
\( A = \left[ -\frac{4}{x} + 2\ln x – \frac{\ln x}{x} \right]_{2}^{4} \)
At \( x = 4 \):
\( -\frac{4}{4} + 2\ln 4 – \frac{\ln 4}{4} = -1 + 2\ln 4 – \frac{\ln 4}{4} \)At \( x = 2 \):
\( -\frac{4}{2} + 2\ln 2 – \frac{\ln 2}{2} = -2 + 2\ln 2 – \frac{\ln 2}{2} \)Subtract:
\( A = \left( -1 + 2\ln 4 – \frac{\ln 4}{4} \right) – \left( -2 + 2\ln 2 – \frac{\ln 2}{2} \right) \)\( = -1 + 2\ln 4 – \frac{\ln 4}{4} + 2 – 2\ln 2 + \frac{\ln 2}{2} \)\( = 1 + 2\ln 4 – 2\ln 2 – \frac{\ln 4}{4} + \frac{\ln 2}{2} \) -
Simplify logarithms:
\( 2\ln 4 – 2\ln 2 = 2\ln\left(\frac{4}{2}\right) = 2\ln 2 \)
\( \ln 4 = 2\ln 2,\) so \( -\frac{\ln 4}{4} = -\frac{2\ln 2}{4} = -\frac{\ln 2}{2} \)Substitute:
\( A = 1 + 2\ln 2 – \frac{\ln 2}{2} + \frac{\ln 2}{2} = 1 + 2\ln 2 \)
Final Answer
\( 1 + 2\ln 2 \)