Question: 1.3 The accumulated response of a system over time is modelled by:
P(x) = \int_0^{x^2} e^t (\sin t + \cos t) \, dt, \quad x \geq 0,
where P(x) represents the total response accumulated from the initial time up to $x^2$. The instantaneous rate at which this accumulated response changes is of interest.
a) Determine the instantaneous rate of change of the accumulated response with respect to $x$ by finding
$P'(x)$. Clearly state the calculus principles used in obtaining your answer.
(4 Marks)
b) Consider the integral $\int e^x(\sin x + \cos x) \, dx$. Evaluate the integral showing all necessary
mathematical steps.
(4 Marks)
1. **Problem Statement:**
We have the accumulated response function:
$$P(x) = \int_0^{x^2} e^t (\sin t + \cos t) \, dt, \quad x \geq 0$$
We want to find:
a) The instantaneous rate of change $P'(x)$.
b) The indefinite integral $\int e^x (\sin x + \cos x) \, dx$.
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2. **Part (a): Find $P'(x)$**
- We use the **Fundamental Theorem of Calculus Part 1** combined with the **Chain Rule**.
- The Fundamental Theorem of Calculus states:
$$\frac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)$$
- Here, $f(t) = e^t (\sin t + \cos t)$ and $g(x) = x^2$.
- Applying the rule:
$$P'(x) = e^{x^2} (\sin x^2 + \cos x^2) \cdot \frac{d}{dx} (x^2)$$
- Compute derivative of $x^2$:
$$\frac{d}{dx} (x^2) = 2x$$
- So,
$$P'(x) = e^{x^2} (\sin x^2 + \cos x^2) \cdot 2x = 2x e^{x^2} (\sin x^2 + \cos x^2)$$
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3. **Part (b): Evaluate $\int e^x (\sin x + \cos x) \, dx$**
- Split the integral:
$$\int e^x (\sin x + \cos x) \, dx = \int e^x \sin x \, dx + \int e^x \cos x \, dx$$
- We use integration by parts or known formulas for these integrals.
- Recall the standard results:
$$\int e^x \sin x \, dx = \frac{e^x}{2} (\sin x - \cos x) + C$$
$$\int e^x \cos x \, dx = \frac{e^x}{2} (\sin x + \cos x) + C$$
- Add them:
$$\int e^x (\sin x + \cos x) \, dx = \frac{e^x}{2} (\sin x - \cos x) + \frac{e^x}{2} (\sin x + \cos x) + C$$
- Simplify inside parentheses:
$$ (\sin x - \cos x) + (\sin x + \cos x) = 2 \sin x$$
- So the integral becomes:
$$\int e^x (\sin x + \cos x) \, dx = \frac{e^x}{2} \cdot 2 \sin x + C = e^x \sin x + C$$
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**Final answers:**
$$\boxed{P'(x) = 2x e^{x^2} (\sin x^2 + \cos x^2)}$$
$$\boxed{\int e^x (\sin x + \cos x) \, dx = e^x \sin x + C}$$