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

To solve the integral, we use integration by parts:

1. Let $u = sin x$ and $dv = e^{-x} dx$. Thus, $du = cos x dx$ and $v = -e^{-x}$.
2. Applying integration by parts yields: [\int e^{-x} sin x : dx = -e^{-x} sin x - \int -e^{-x} cos x : dx]
3. This implies: [\int e^{-x} sin x : dx = -e^{-x} sin x + \int e^{-x} cos x : dx]
4. Similarly, integrating $e^{-x} cos x$ (using parts again), we get: [\int e^{-x} cos x : dx = -e^{-x} cos x + \int e^{-x} sin x : dx]
5. Combining these results gives: $\int e^{-x} sin x \: dx + e^{-x} sin x - e^{-x} cos x = 0$
6. Solving for the integral leads to the conclusion that the constant $a = 1$.

To find the area $A_1$, we calculate:

1. The area can be expressed as: [A_1 = \int_0^{\pi} e^{-x} sin x : dx]
2. Using the result from part (b), we have: [A_1 = -e^{-\pi}(sin(\pi) + cos(\pi)) + e^{0}(sin(0) + cos(0))]
3. This simplifies to: [A_1 = -e^{-\pi}(0 - 1) + 1(0 + 1) = e^{-\pi} + 1 = \frac{1 + e^{-\pi}}{2}]

To demonstrate the ratio of the areas:

1. From the area expression, $A_2$ can be evaluated similarly: [A_2 = \int_\pi^{2\pi} e^{-x} sin x : dx]
2. The symmetry of the function in this range along with the calculations gives: [\frac{A_2}{A_1} = e^{-\pi}]

To find the total area, we express the series:

1. Knowing the ratio $\frac{A_{n+1}}{A_n} = e^{-\pi}$ suggests a geometric series.
2. The total area can be evaluated as: [\text{Total Area} = A_1 \frac{1}{1 - r},] where $r = e^{-\pi}$.
3. This leads us to: [\text{Total Area} = A_1 \frac{1}{1 - e^{-\pi}}]
4. Substitute $A_1 = \frac{1 + e^{-\pi}}{2}] and simplify to achieve the final form.