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:
\[
\boxed{3e^4 + 1}
\]