Edexcel IAL January 2023 P4 (WMA14/01) Q4 Integration by Substitution, Integration by Parts
(a) Using the substitution \( u = \sqrt{2x + 1} \), show that
\(\int_{4}^{12} \sqrt{8x + 4} e^{\sqrt{2x+1}} dx\)
may be expressed in the form
\(\int_{a}^{b} ku^2 e^{u} du\)
where a, b and k are constants to be found.(4)
(b) Hence find, by algebraic integration, the exact value of
\(\int_{4}^{12} \sqrt{8x + 4} e^{\sqrt{2x+1}} dx\)
giving your answer in simplest form.(5)
Solution to Part a: Integration by Substitution
Key Concepts Used:
- Substitution Method: Let \( u = \sqrt{2x + 1} \) to simplify the integral.
- Change of Limits: Convert x-limits to u-limits.
- Algebraic Manipulation: Express \(\sqrt{8x + 4}\) in terms of \( u \).
- Differential Relation: Find \( dx \) in terms of \( du \).
Step-by-Step Solution:
We are given: \(I = \int_{4}^{12} \sqrt{8x + 4} \cdot e^{\sqrt{2x+1}} dx\)
-
Simplify the Square Root:
\(\sqrt{8x + 4} = \sqrt{4(2x + 1)} = 2\sqrt{2x + 1}\)
So:
\(I = \int_{4}^{12} 2\sqrt{2x+1} \cdot e^{\sqrt{2x+1}} dx\) -
Substitution: Let:
\( u = \sqrt{2x+1} \quad \Rightarrow \quad u^2 = 2x + 1 \)
Differentiate implicitly:
\( 2u du = 2 dx \quad \Rightarrow \quad dx = u du \) -
Change Limits:
When \( x = 4 \): \( u = \sqrt{2(4) + 1} = \sqrt{9} = 3 \)When \( x = 12 \): \( u = \sqrt{2(12) + 1} = \sqrt{25} = 5 \)
-
Rewrite the Integral: Substitute \(\sqrt{2x + 1} = u\) and \(dx = u du\):
\( I = \int_{3}^{5} 2u \cdot e^{u} \cdot u du = \int_{3}^{5} 2u^2 e^{u} du \)
So, the integral is expressed as:
\( I = \int_{3}^{5} 2u^2 e^{u} du \)Thus, \(a = 3\), \(b = 5\), \(k = 2\).
Final Answer
\( \int_{3}^{5} 2u^2 e^{u} du \)
Solution to Part b: Integration by Parts
Key Concepts Used:
- Integration by Parts: \( \int u dv = uv – \int v du \).
- Repeated Application: Apply integration by parts twice to reduce the power of \( u \).
- Exponential Integration: \( \int e^u du = e^u + C \).
- Definite Integration: Evaluate the antiderivative at the bounds.
Step-by-Step Solution:
We need to evaluate: \(I = 2 \int_3^5 u^2 e^u du\)
-
First Integration by Parts:
Let:
\( u = u^2 \rightarrow du = 2u du \)
\( dv = e^u du \rightarrow v = e^u \)Apply:
\(\int u^2 e^u du = u^2 e^u – \int 2ue^u du = u^2 e^u – 2 \int ue^u du\) -
Second Integration by Parts for \( \int ue^u du \): Let:
\( u = u \rightarrow du = du \)
\( dv = e^u du \rightarrow v = e^u \)Apply:
\(\int ue^u du = ue^u – \int e^u du = ue^u – e^u\) -
Combine Results:
\(\int u^2 e^u du = u^2 e^u – 2(ue^u – e^u) = u^2 e^u – 2ue^u + 2e^u\)
-
Now Evaluate from 3 to 5:
\([u^2 e^u – 2ue^u + 2e^u]_{3}^{5}\)At \( u = 5 \): \(25e^5 – 10e^5 + 2e^5 = 17e^5\)At \( u = 3 \): \(9e^3 – 6e^3 + 2e^3 = 5e^3\)
So:
\(\int_{3}^{5} u^2 e^u du = 17e^5 – 5e^3\)Multiply by 2:
\(I = 2(17e^5 – 5e^3) = 34e^5 – 10e^3\)
Final Answer
\(34e^5 – 10e^3\)