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 5 - 2006 - Paper 7

Let ( u = x^2 ) and ( dv = e^{2x} dx ).

Then, differentiate & integrate to find:

[ du = 2x dx \quad v = \frac{e^{2x}}{2} ]

Using integration by parts:

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

Substituting, we have:

$= \left[ x^2 \cdot \frac{e^{2x}}{2} \right] - \int \frac{e^{2x}}{2} (2x) \, dx$.

This simplification leads to:

$= \left[ \frac{x^2 e^{2x}}{2} - \int xe^{2x} dx \right].$

The remaining integral $\int xe^{2x} dx$ requires integration by parts again, choosing:

( u = x, , dv = e^{2x} dx ).

Following the integration process again yields:

$\int xe^{2x} dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4}.$

Combining results will give the final form of the integral in Step 2.

Now substitute the upper limit of 3 and lower limit of 1 into the resulting expression:

Evaluate:

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

This provides the volume as:

$V = \pi \left( \text{Evaluate the above expression} \right).$