Edexcel IAL January 2021 P4 (WMA14/01) Q7 Integration by Parts, Area under Curves
(a) Find \(\int e^{2x} \sin x dx\)
Figure 2 shows a sketch of part of the curve with equation
\( y = e^{2x} \sin x \quad x \geq 0 \)
The finite region \(R\) is bounded by the curve and the x-axis and is shown shaded in Figure 2.
(b) Show that the exact area of \(R\) is
\(\frac{e^{2\pi+1}}{5}\)
(Solutions relying on calculator technology are not acceptable.)(2)
Solution to Part a: Integration by Parts
Key Concepts Used:
- Integration by Parts Formula: \(\int u dv = uv – \int v du\)
- Cyclic Integration: Applying integration by parts twice to handle the product of exponential and trigonometric functions.
- Solving for the Integral: Combining like terms to isolate the desired integral.
- Exponential and Trigonometric Integrals: Knowledge of derivatives and integrals of \(e^{ax}\) and \(\sin bx\), \(\cos bx\).
Step-by-Step Solution:
We want to find: \(I = \int e^{2x} \sin x dx\)
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First Integration by Parts: Let:
\(u = e^{2x} \rightarrow du = 2e^{2x} dx\)
\(dv = \sin x dx \rightarrow v = -\cos x\)Apply integration by parts:
\(I = e^{2x} (-\cos x) – \int (-\cos x) \cdot 2e^{2x} dx = -e^{2x} \cos x + 2 \int e^{2x} \cos x dx\) -
Second Integration by Parts for \(\int e^{2x} \cos x dx\): Let:
\(u = e^{2x} \rightarrow du = 2e^{2x} dx\)
\(dv = \cos x dx \rightarrow v = \sin x\)Apply:
\(\int e^{2x} \cos x dx = e^{2x} \sin x – \int \sin x \cdot 2e^{2x} dx = e^{2x} \sin x – 2 \int e^{2x} \sin x dx\)Notice that \(\int e^{2x} \sin x dx = I\), so:
\(\int e^{2x} \cos x dx = e^{2x} \sin x – 2I\) -
Substitute Back into \(I\):
\(I = -e^{2x} \cos x + 2(e^{2x} \sin x – 2I) = -e^{2x} \cos x + 2e^{2x} \sin x – 4I\)
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Solve for \(I\):
\(I + 4I = -e^{2x}\cos x + 2e^{2x}\sin x\)
\(5I = e^{2x}(2\sin x – \cos x)\)
\(I = \frac{1}{5}e^{2x}(2\sin x – \cos x) + C\)
Final Answer
\(\frac{1}{5}e^{2x}(2\sin x – \cos x) + C\)
Solution to Part b: Area of Region R
Key Concepts Used:
- Definite Integral for Area: The area under \( y = e^{2x} \sin x \) from \( x = 0 \) to \( x = \pi \).
- Evaluation of Antiderivative: Use the result from part (a) to compute the definite integral.
- Trigonometric Values: \(\sin \pi = 0\), \(\cos \pi = -1\), \(\sin 0 = 0\), \(\cos 0 = 1\).
Step-by-Step Solution:
The area \( A \) of region R is: \(A = \int_{0}^{\pi} e^{2x} \sin x dx\)
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From part (a), the antiderivative is:
\(F(x) = \frac{1}{5}e^{2x}(2\sin x – \cos x)\)
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Evaluate from 0 to \(\pi\):
At \( x = \pi \): \(F(\pi) = \frac{1}{5}e^{2\pi}(2\sin \pi – \cos \pi) = \frac{1}{5}e^{2\pi}(0 – (-1)) = \frac{1}{5}e^{2\pi}\)At \( x = 0 \): \(F(0) = \frac{1}{5}e^{0}(2\sin 0 – \cos 0) = \frac{1}{5}(0 – 1) = -\frac{1}{5}\)
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Compute the Definite Integral:
\(A = F(\pi) – F(0) = \frac{1}{5} e^{2\pi} – \left(-\frac{1}{5}\right) = \frac{1}{5} e^{2\pi} + \frac{1}{5} = \frac{1}{5} (e^{2\pi} + 1)\)
Final Answer
\(\frac{e^{2\pi} + 1}{5}\)