Figure 1 shows the finite region R, which is bounded by the curve $y = xe^x$, the line $x = 1$, the line $x = 3$ and the x-axis - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 7

Using integration by parts, let:

• $u = x^2$, hence $du = 2x \, dx$
• $dv = e^{2x} \, dx$, hence $v = \frac{1}{2} e^{2x}$.

Now apply the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Substituting:

$\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x \, dx$

This simplifies to:

$= \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx$

Now, we need to evaluate the integral $\int x e^{2x} \, dx$ again using integration by parts:

Let:

• $u = x$, hence $du = dx$
• $dv = e^{2x} \, dx$, hence $v = \frac{1}{2} e^{2x}$.

Then,

$\int x e^{2x} \; dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \; dx$

The integral of $e^{2x}$ is:

$= \frac{1}{2} e^{2x}$

So combining the results:

$\int x e^{2x} \; dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}$

Putting this back, we find that:$

V = \pi \left[ \frac{1}{2} x^2 e^{2x} - (\frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}) \right]_{1}^{3}$