Edexcel IAL October 2023 P4 (WMA14/01) Q3 Integration by Substitution, Integration by Parts
In this question you must show all stages of your working.
Solutions based on calculator technology are not acceptable.
Solutions based on calculator technology are not acceptable.
i- Use integration by parts to find the exact value of
\(\int_{2}^{4} x^2 e^{2x} dx\)
giving your answer in simplest form.(5)
ii- Use integration by substitution to show that
\(\int_{3}^{\frac{21}{2}} \frac{4x}{(2x – 1)^2} dx = a + \ln b\)
where \(a\) and \(b\) are constants to be found.(7)
Solution to Part i: Integration by Parts
Key Concepts Used:
- Integration by Parts Formula: \(\int u dv = uv – \int v du\)
- Repeated Application: Applying integration by parts twice to reduce the power of \(x\)
- Exponential Integration: \(\int e^{ax} dx = \frac{1}{a} e^{ax} + C\)
- Definite Integration: Evaluating the antiderivative at bounds
Step-by-Step Solution:
We want to evaluate: \(I = \int_{0}^{4} x^2 e^{2x} dx\)
-
First Integration by Parts: Let:
\( u = x^2 \rightarrow du = 2x dx \)\( dv = e^{2x} dx \rightarrow v = \frac{1}{2} e^{2x} \)
Apply integration by parts:
\( I = \left[ \frac{1}{2} x^2 e^{2x} \right]_0^4 – \int_0^4 \frac{1}{2} e^{2x} \cdot 2x dx = \left[ \frac{1}{2} x^2 e^{2x} \right]_0^4 – \int_0^4 x e^{2x} dx \) -
Second Integration by Parts for \(\int xe^{2x} dx\): let:
\( u = x \rightarrow du = dx \)\( dv = e^{2x} dx \rightarrow v = \frac{1}{2} e^{2x} \)
Apply:
\( \int xe^{2x} dx = \frac{1}{2} xe^{2x} – \int \frac{1}{2} e^{2x} dx = \frac{1}{2} xe^{2x} – \frac{1}{4} e^{2x} \) -
Substitute Back into \(I\):
\( I = \left[ \frac{1}{2} x^2 e^{2x} \right]_0^4 – \left[ \frac{1}{2} xe^{2x} – \frac{1}{4} e^{2x} \right]_0^4 \)
-
Evaluate at Bounds:
At \( x = 4 \):
\( \frac{1}{2} (16) e^8 = 8e^8 \)
\( \frac{1}{2} (4) e^8 – \frac{1}{4} e^8 = 2e^8 – \frac{1}{4} e^8 = \frac{7}{4} e^8 \)At \( x = 0 \):
\( \frac{1}{2} (0)^2 e^0 = 0 \)
\( \frac{1}{2} (0) e^0 – \frac{1}{4} e^0 = 0 – \frac{1}{4} = -\frac{1}{4} \)So:
\( I = (8e^8 – 0) – \left( \frac{7}{4} e^8 – \left( -\frac{1}{4} \right) \right) = 8e^8 – \left( \frac{7}{4} e^8 + \frac{1}{4} \right) \)\( = 8e^8 – \frac{7}{4} e^8 – \frac{1}{4} = \left( \frac{32}{4} e^8 – \frac{7}{4} e^8 \right) – \frac{1}{4} = \frac{25}{4} e^8 – \frac{1}{4} \)
Final Answer
\( \frac{25}{4} e^8 – \frac{1}{4} \)
Solution to Part ii: Integration by Substitution
Key Concepts Used:
- Substitution Method: Let \( u = 2x – 1 \) to simplify the denominator.
- Change of Limits: Convert \( x \)-limits to \( u \)-limits.
- Logarithmic Integration: \( \int \frac{1}{u} du = \ln |u| + C \).
- Algebraic Manipulation: Express \( x \) in terms of \( u \) and simplify the integrand.
Step-by-Step Solution:
We want to show: \(\int_{3}^{21/2} \frac{4x}{(2x – 1)^2} dx = a + \ln b\)
-
Substitution: Let:
\( u = 2x – 1 \Rightarrow du = 2 dx \Rightarrow dx = \frac{du}{2} \)
Also, solve for \( x \):
\( u = 2x – 1 \Rightarrow 2x = u + 1 \Rightarrow x = \frac{u + 1}{2} \) -
Change of Limits:
When \( x = 3 \): \( u = 2(3) – 1 = 6 – 1 = 5 \)When \( x = 21/2 \): \( u = 2\left(\frac{21}{2}\right) – 1 = 21 – 1 = 20 \)
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Rewrite the Integral: Substitute \( x = \frac{u+1}{2} \) and \( dx = \frac{du}{2} \):
\(\int_{x=3}^{x=21/2} \frac{4x}{(2x-1)^2} dx = \int_{u=5}^{u=20} \frac{u+1}{u^2} du\)
Simplify:
\(= \int_5^{20} \frac{2(u+1)}{u^2} \cdot \frac{1}{2} du = \int_5^{20} \frac{u+1}{u^2} du\)\(= \int_5^{20} \left( \frac{1}{u} + \frac{1}{u^2} \right) du = \int_5^{20} u^{-1} du + \int_5^{20} u^{-2} du\) -
Integrate:
\(= [\ln|u||_5^{20} + \left[ -\frac{1}{u} \right]_5^{20} = (\ln 20 – \ln 5) + \left( -\frac{1}{20} + \frac{1}{5} \right)\)
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Simplify:
\(\ln 20 – \ln 5 = \ln \left( \frac{20}{5} \right) = \ln 4\)\(- \frac{1}{20} + \frac{1}{5} = – \frac{1}{20} + \frac{4}{20} = \frac{3}{20}\)
So:
\(\int_3^{21/2} \frac{4x}{(2x-1)^2} dx = \ln 4 + \frac{3}{20}\)Thus, \( a = \frac{3}{20}, b = 4 \).
Final Answer
\( \frac{3}{20} + \ln 4 \)