Cambridge Additional Mathematics 0606 : 2023 October-November Paper 23 Q9 (Solution)

Question 9(a)

Find $\frac{dy}{dx}$

Steps of Solution

Given $y = xe^{2x}$ :

$\frac{dy}{dx} = x \cdot \frac{d}{dx}(e^{2x}) + e^{2x} \cdot \frac{d}{dx}(x)$
$\frac{dy}{dx} = x \cdot 2e^{2x} + e^{2x} \cdot 1$
$\frac{dy}{dx} = 2xe^{2x} + e^{2x}$
$\frac{dy}{dx} = e^{2x}(2x + 1)$

Explanation of the Steps

Given the function $y = xe^{2x}$ , we need to find its derivative with respect to $x$ . We’ll use the product rule, which states that for two functions $u(x)$ and $v(x)$ , the derivative of their product is:

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

Here, $u = x$ and $v = e^{2x}$ .

First, we find the derivatives of $u$ and $v$ :
$\frac{du}{dx} = 1$
$\frac{dv}{dx} = \frac{d}{dx}(e^{2x}) = 2e^{2x}$

Applying the product rule:
$\frac{dy}{dx} = x \cdot \frac{d}{dx}(e^{2x}) + e^{2x} \cdot \frac{d}{dx}(x)$
$\frac{dy}{dx} = x \cdot 2e^{2x} + e^{2x} \cdot 1$
$\frac{dy}{dx} = 2xe^{2x} + e^{2x}$

Factor out $e^{2x}$ :
$\frac{dy}{dx} = e^{2x}(2x + 1)$

Question 9(b)

Find the equation of the normal to the curve at $x = 1$

Steps of Solution

Given the curve $y = xe^{2x}$ :
$\left.\frac{dy}{dx}\right|_{x=1} = e^{2(1)}(2(1) + 1)$
$\left.\frac{dy}{dx}\right|_{x=1} = e^2(2 + 1)$
$\left.\frac{dy}{dx}\right|_{x=1} = 3e^2$

The slope of the normal line is:
$m_{\text{normal}} = -\frac{1}{3e^2}$

The point on the curve at $x = 1$ :
$y = 1 \cdot e^{2(1)} = e^2$

Equation of the normal line using the point-slope form $y – y_1 = m(x – x_1)$ :
$y – e^2 = -\frac{1}{3e^2}(x – 1)$

Explanation of the Steps

Given the curve $y = xe^{2x}$ , we first need to find the derivative $\frac{dy}{dx}$ at $x = 1$ . We already determined that:
$\frac{dy}{dx} = e^{2x}(2x + 1)$

Evaluating at $x = 1$ :
$\left.\frac{dy}{dx}\right|_{x=1} = e^{2(1)}(2(1) + 1)$
$\left.\frac{dy}{dx}\right|_{x=1} = e^2(2 + 1)$
$\left.\frac{dy}{dx}\right|_{x=1} = 3e^2$

The gradient of the tangent to the curve at $x = 1$ is $3e^2$ . The gradient of the normal is the negative reciprocal of the gradient of the tangent:
$m_{\text{normal}} = -\frac{1}{3e^2}$

Next, we find the coordinates of the point on the curve at $x = 1$ :
$y = xe^{2x}$
$y = 1 \cdot e^{2(1)} = e^2$

So the point is $(1, e^2)$ . The equation of the normal line in point-slope form $y – y_1 = m(x – x_1)$ is:
$y – e^2 = -\frac{1}{3e^2}(x – 1)$

Simplifying this equation gives us the equation of the normal line.

Question 9(c)

Use your answer to part (a) to find the exact value of $\int_{0}^{2} 2xe^{2x} \, dx$

Steps of Solution

Given:
$\int 2xe^{2x} \, dx$

Using integration by parts where $u = x$ and $dv = e^{2x}dx$ :
$du = dx$
$v = \frac{e^{2x}}{2}$

So:
$\int 2xe^{2x} \, dx = 2 \left( x \cdot \frac{e^{2x}}{2} – \int \frac{e^{2x}}{2} \, dx \right)$
$= xe^{2x} – \int e^{2x} \, dx$
$= xe^{2x} – \frac{e^{2x}}{2} + C$

Evaluating from 0 to 2:
$\left[ xe^{2x} – \frac{e^{2x}}{2} \right]_0^2$
$= \left( 2e^{4} – \frac{e^{4}}{2} \right) – \left( 0 – \frac{e^{0}}{2} \right)$
$= 2e^4 – \frac{e^4}{2} + \frac{1}{2}$
$= \frac{4e^4}{2} – \frac{e^4}{2} + \frac{1}{2}$
$= \frac{3e^4}{2} + \frac{1}{2}$
$= \frac{3e^4 + 1}{2}$

Explanation of the Steps

To find the exact value of the integral $\int_{0}^{2} 2xe^{2x} \, dx$ , we use the method of integration by parts. Let:
$u = x \quad \text{and} \quad dv = 2e^{2x}dx$

Then:
$du = dx \quad \text{and} \quad v = \int 2e^{2x}dx = e^{2x}$

Using the integration by parts formula $\int u \, dv = uv – \int v \, du$ , we get:
$\int 2xe^{2x} \, dx = 2 \left( x \cdot e^{2x} – \int e^{2x} \, dx \right)$

Solving the integral $\int e^{2x} \, dx$ :
$\int e^{2x} \, dx = \frac{e^{2x}}{2}$

So:
$= 2 \left( x \cdot e^{2x} – \frac{e^{2x}}{2} \right)$
$= 2xe^{2x} – e^{2x} + C$

Evaluating the definite integral from 0 to 2:
$\left[ 2xe^{2x} – e^{2x} \right]_0^2$
$= \left( 2(2)e^{2(2)} – e^{2(2)} \right) – \left( 2(0)e^{2(0)} – e^{2(0)} \right)$
$= \left( 4e^{4} – e^{4} \right) – \left( 0 – 1 \right)$
$= 3e^4 + 1$

Thus, the exact value of the integral is:
$\boxed{3e^4 + 1}$

Question 9(a)

Steps of Solution

Explanation of the Steps

Question 9(b)

Steps of Solution

Explanation of the Steps

Question 9(c)

Steps of Solution

Explanation of the Steps

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