Evaluate ∫ from 0 to 2 of x e^x dx Step by Step

Step by step

The question

Question: find the integral from 0 to 2 of x eˣ.

∫ from 0 to 2 of xe^x dx =

1 Choose u and dv

Choose u = x: its derivative is simpler, just a number.

dv is the rest: eˣ dx, which we can integrate.

u = x, dv = e^x dx

2 Find du and v

du: the derivative of u, times dx.

v: the integral of dv.

e to the x stays e to the x.

du = 1 dx, v = e^x

3 Use the rule

The rule: the integral of u dv is u v, minus the integral of v du.

It comes from the product rule, backwards.

∫ u dv = uv − ∫ v du

u times v, minus the integral of v times du.

= xe^x − ∫ e^x dx

4 The new integral

The new integral is easy: just a number times e to a power.

= xe^x − e^x

Careful!

It is u v minus the integral of v du.

Not plus!

Careful: not uv + ∫ v du ✗

Open the bracket.

No + C: top minus bottom.

= [ xe^x − e^x ] from 0 to 2

5 Top minus bottom

Put in 2, then take away the same with 0.

= (2 × e² − e²) − (0 × 1 − 1)

e to the 0 is 1.

e⁰ = 1

Tidy up: the integral is e² + 1, about 8.389.

= e² + 1 ≈ 8.389

6 Check

Let's check with a calculator: it adds up 1000 thin strips.

That gives about 8.389, the same as our answer.

Right!

1000 thin strips: ≈ 8.389

Let's look back.

First: choose u and dv.

Then: find du and v.

Then: use the rule.

Then: the new integral.

Then: top minus bottom.

Last: check.

Answer