Let's tackle each part of the problem step by step.

(i) Show that $\frac{dy}{dx} = 2e^{2x}(kx - 1)$, where $k$ is a constant.

The given function is:

$y = 2(x - 1)e^{2x}$

To find $\frac{dy}{dx}$, we apply the product rule, which states:

$\frac{d}{dx}[u \cdot v] = u'v + uv'$

Here:

• $u = 2(x - 1)$
• $v = e^{2x}$

Step 1: Differentiate $u$:

$u = 2(x - 1) \implies u' = 2$

Step 2: Differentiate $v$:

$v = e^{2x} \implies v' = 2e^{2x} \quad \text{(by chain rule)}$

Step 3: Apply the product rule:

$\frac{dy}{dx} = u'v + uv'$ $\frac{dy}{dx} = 2e^{2x} + 2(x - 1)(2e^{2x})$

Simplify: $\frac{dy}{dx} = 2e^{2x} + 4(x - 1)e^{2x}$

Factorize $2e^{2x}$: $\frac{dy}{dx} = 2e^{2x}[1 + 2(x - 1)]$ $\frac{dy}{dx} = 2e^{2x}(2x - 1)$

Here, $k = 2$. Thus: $\frac{dy}{dx} = 2e^{2x}(kx - 1)$

(ii) Use your answer to part (i) to find the exact value of $\int_0^1 xe^{2x} \, dx$.

From (i), we know: $\frac{dy}{dx} = 2e^{2x}(2x - 1)$

Rewriting $2x e^{2x}$ from this expression: $2x e^{2x} = \frac{dy}{dx} + 2e^{2x}$

Thus: $\int_0^1 x e^{2x} \, dx = \frac{1}{2} \int_0^1 \left( \frac{dy}{dx} + 2e^{2x} \right) dx$

Split into two integrals: $\int_0^1 x e^{2x} \, dx = \frac{1}{2} \left[ \int_0^1 \frac{dy}{dx} \, dx + \int_0^1 2e^{2x} \, dx \right]$

1. First Integral: $\int_0^1 \frac{dy}{dx} \, dx = y(1) - y(0)$ $y = 2(x - 1)e^{2x} \quad \text{so:}$ $y(1) = 2(1 - 1)e^2 = 0, \quad y(0) = 2(0 - 1)e^0 = -2$ $\int_0^1 \frac{dy}{dx} \, dx = 0 - (-2) = 2$

2. Second Integral: $\int_0^1 2e^{2x} \, dx = 2 \int_0^1 e^{2x} \, dx$ The integral of $e^{2x}$ is: $\int e^{2x} \, dx = \frac{1}{2} e^{2x}$ Evaluate from 0 to 1: [ 2 \int_0^1 e^{2x} , dx = 2 \left[ \frac{1}{2}e^{2x} \right]_0^1 = \left[ e^2 - 1 \right] ]

Thus: $\int_0^1 x e^{2x} \, dx = \frac{1}{2} \left[ 2 + (e^2 - 1) \right]$ $\int_0^1 x e^{2x} \, dx = \frac{1}{2}(e^2 + 1)$

(iii) Find the rate of change of $x$ when $x = 2$, given that $y$ is changing at a constant rate of $2e$ units per second.

We are given: $\frac{dy}{dt} = 2e, \quad x = 2$

From (i): $\frac{dy}{dx} = 2e^{2x}(2x - 1)$

When $x = 2$: $\frac{dy}{dx} = 2e^{4}(2(2) - 1) = 2e^{4}(4 - 1) = 6e^{4}$

The chain rule relates $\frac{dy}{dt}$, $\frac{dy}{dx}$, and $\frac{dx}{dt}$: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

Solve for $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{\frac{dy}{dt}}{\frac{dy}{dx}}$ $\frac{dx}{dt} = \frac{2e}{6e^4} = \frac{1}{3e^3}$

Final Answers:

(i) $\frac{dy}{dx} = 2e^{2x}(2x - 1)$
(ii) $\int_0^1 x e^{2x} \, dx = \frac{1}{2}(e^2 + 1)$
(iii) $\frac{dx}{dt} = \frac{1}{3e^3}$

Would you like me to clarify any part of this? Here are five related questions for deeper understanding:

1. How does the product rule simplify the differentiation in (i)?
2. What role does substitution play in solving integrals like in (ii)?
3. Why is the chain rule necessary for finding rates of change in (iii)?
4. How can we interpret the constant $k$ in (i) in other contexts?
5. Can $e^{2x}$ integrals be generalized for other functions?

Tip: Always verify your integral limits when evaluating definite integrals to avoid errors.