16 (a) y = e^{-x} (sin x + cos x) Find $ \frac{dy}{dx} $ Simplify your answer. (3 marks) 16 (b) Hence, show that $ \int e^{-x} sin x dx = a e^{-x} (sin x + cos x) + c $ where $ a $ is a rational number. (2 marks) 16 (c) A sketch of the graph of $ y = e^{-x} sin x $ for $ x \geq 0 $ is shown below. The areas of the finite regions bounded by the curve and the x-axis are denoted by $ A_1, A_2, ..., A_n, ... $ (i) Find the exact value of the area $ A_1 $ (3 marks) (ii) Show that $ \frac{A_2}{A_1} = e^{-\pi} $ (4 marks) (iii) Given that $ \frac{A_{n+1}}{A_n} = e^{-\pi} $ show that the exact value of the total area enclosed between the curve and the x-axis is $ \frac{1 + e^{-\pi}}{2(e - 1)} $ (4 marks)

A-Level AQA Maths Pure Further Integration: 16 (a) y = e^{-x} (sin x + co

16 (a) y = e^{-x} (sin x + cos x) Find $ \frac{dy}{dx} $ Simplify your answer - AQA - A-Level Maths Pure - Question 16 - 2019 - Paper 1

Question 16

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16 (a) y = e^{-x} (sin x + cos x) Find $ \frac{dy}{dx} $ Simplify your answer. (3 marks) 16 (b) Hence, show that \( \int e^{-x} sin x dx = a e^{-x} (sin... show full transcript

Worked Solution & Example Answer:16 (a) y = e^{-x} (sin x + cos x) Find $ \frac{dy}{dx} $ Simplify your answer - AQA - A-Level Maths Pure - Question 16 - 2019 - Paper 1

Step 1

Find $ \frac{dy}{dx} $

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Answer

To find ( \frac{dy}{dx} ), we will use the product rule. Let ( u = e^{-x} ) and ( v = sin x + cos x ). Then, the product rule states that:

[ \frac{d(uv)}{dx} = u'v + uv' ]

Calculating each component: [ u' = rac{d}{dx}(sin x + cos x) = cos x - sin x ] [ u' = -e^{-x} ]

Substituting back, we get: [ \frac{dy}{dx} = e^{-x}(cos x + cos x) + (sin x + cos x)(-e^{-x}) = e^{-x}(2cos x - (sin x + cos x)) ]

Simplifying this, we have:

[ \frac{dy}{dx} = e^{-x}(-sin x + cos x) ]

Step 2

Hence, show that $\int e^{-x} sin x dx = a e^{-x} (sin x + cos x) + c$

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Answer

To show this, we can apply integration by parts. Let ( u = sin x ) and ( dv = e^{-x}dx ). Therefore, we have:

[ \int e^{-x} sin x dx = -e^{-x} (sin x - cos x) + \int e^{-x} cos x dx ]

By applying integration by parts again, we find: [ \int e^{-x} cos x dx = e^{-x} (cos x + sin x) + c ]

Thus, combining these results, we can express: [ \int e^{-x} sin x dx = a e^{-x} (sin x + cos x) + c ] for some rational number ( a ).

Step 3

Find the exact value of the area $ A_1 $

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Answer

The area ( A_1 ) can be found using the integral:

[ A_1 = \int_0^{\pi} e^{-x} sin x dx = -e^{-x} (sin x + cos x) \big|_0^{\pi} = -e^{-\pi} (-1) - (0 + 1) ]

Evaluating this provides: [ A_1 = (e^{-\pi} - 1) ]

Step 4

Show that $ \frac{A_2}{A_1} = e^{-\pi} $

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Answer

Using the ratio of areas:

[ A_2 = e^{-\pi} A_1 ] implies that: [ \frac{A_2}{A_1} = e^{-\pi} ] as desired.

Step 5

Show that the exact value of the total area enclosed is $ \frac{1 + e^{-\pi}}{2(e - 1)} $

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Answer

Using the geometric series based on the ratio: [ \sum_{n=0}^{\infty} A_1 (e^{-\pi})^n ]

This converges to: [ \frac{A_1}{1 - e^{-\pi}} = \frac{e^{-\pi} - 1}{2(e - 1)} ]

Collectively yielding: [ \frac{1 + e^{-\pi}}{2(e - 1)} ] for the total area.

16 (a) y = e^{-x} (sin x + cos x) Find \( \frac{dy}{dx} \) Simplify your answer - AQA - A-Level Maths Pure - Question 16 - 2019 - Paper 1

Worked Solution & Example Answer:16 (a) y = e^{-x} (sin x + cos x) Find \( \frac{dy}{dx} \) Simplify your answer - AQA - A-Level Maths Pure - Question 16 - 2019 - Paper 1

Find \( \frac{dy}{dx} \)

Hence, show that \(\int e^{-x} sin x dx = a e^{-x} (sin x + cos x) + c\)

Find the exact value of the area \( A_1 \)

Show that \( \frac{A_2}{A_1} = e^{-\pi} \)

Show that the exact value of the total area enclosed is \( \frac{1 + e^{-\pi}}{2(e - 1)} \)

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