Integration by parts

Warm-up

Three quick questions from earlier lessons.

Discover

The product rule says $(x e^x)' = e^x + x e^x$ . Integrate both sides: $x e^x = \displaystyle\int e^x\,dx + \int x e^x\,dx$ . So what is $\displaystyle\int x e^x\,dx$ ?

$x e^x - e^x + C$
$\dfrac{x^2}{2}e^x + C$

Key idea

Trade one integral for an easier one.

Integration by parts:

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

. For definite integrals:

\displaystyle\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du

. Choose

u

so that

du

is simpler.

Worked example

$\displaystyle\int x\sin x\,dx$ .

u = x

du = dx

v = -\cos x

u = x simplifies; dv = sin x dx is easy to integrate.

-x\cos x + \displaystyle\int \cos x\,dx

uv − ∫v du.

-x\cos x + \sin x + C

Finish.

Your turn, guided

Your turn: $\displaystyle\int x e^{2x}\,dx$ , $u = x$ , $v = \tfrac{1}{2}e^{2x}$ .

\tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + C

½xe^{2x} − ∫½e^{2x} dx.

Practice ★

Find $F$ with $F'(x) = 8x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 6x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 7x e^x$ and $F(0) = 0$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_0^{\pi/2} 8x\cos x\,dx$ .
Evaluate $\displaystyle\int_0^{\pi/2} 3x\cos x\,dx$ .

Evaluate $\displaystyle\int_1^{e^{ 3 } } \ln x\,dx$ .

Boss problem. Evaluate $\displaystyle\int_0^{ 2 } x e^x\,dx$ by parts.

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = 8x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 6x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 7x e^x$ and $F(0) = 0$ .

Summary card

Integration by parts

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

Watch out: the minus sign, and signs inside $\int v\,du$ .

End of the lesson

Next lesson: Integration by parts: choosing u, repeated parts Back to the chapter

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