Edexcel Pure Paper 2 2024 Question 11 Solution

We’re finding the area, so we’re dealing with:

\int_0^1 8x^2 e^{-3x} , dx

Start with the usual split.

Take the polynomial as the part to differentiate — that reduces nicely. The exponential stays manageable when integrated.

So:

Differentiate 8x^2 → 16x
Integrate e^{-3x} → -\frac{1}{3}e^{-3x}

That gives:

= -\frac{8x^2}{3}e^{-3x} + \int \frac{16x}{3} e^{-3x} dx

Still an x there, so it doesn’t stop yet.

Same idea again.

Now treat \frac{16x}{3} as the part to differentiate → becomes \frac{16}{3}
And again, integrating e^{-3x} gives -\frac{1}{3}e^{-3x}

So this next bit turns into:

= -\frac{16x}{9}e^{-3x} + \int \frac{16}{9} e^{-3x} dx

At this point it finally settles down — no x left.

That last integral:

= -\frac{16}{27}e^{-3x}

So altogether:

\int 8x^2 e^{-3x} dx = -\frac{8x^2}{3}e^{-3x} – \frac{16x}{9}e^{-3x} – \frac{16}{27}e^{-3x}

Now deal with the limits.

Start with x = 1.

You get three terms, all with e^{-3}. It’s easier just to combine them straight away rather than keeping them separate.

So:

-\frac{8}{3} – \frac{16}{9} – \frac{16}{27}

Put them over 27:

\frac{72}{27} + \frac{48}{27} + \frac{16}{27} = \frac{136}{27}

So the whole thing is:

-\frac{136}{27}e^{-3}

Now the lower limit.

At x = 0, the terms with x just disappear. What’s left is:

-\frac{16}{27}

(since e^0 = 1)

Now subtract.

It’s easy to rush this bit, but it matters.

Upper minus lower:

-\frac{136}{27}e^{-3} – (-\frac{16}{27})

So the final answer comes out as:

\frac{16}{27} – \frac{136}{27}e^{-3}