Integration by Parts

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Integration by Parts

Derivation of Integration by Parts

Recall the product rule:

(uv)' = u' v + uv'

uv' = (uv)' - u' v

Integrating both sides, we have that

$ \int uv' dx = \int (uv)' dx - \int u' v dx $

$ = uv - \int u' v dx $

Theorem: Integration by Parts

Let u and v be differentiable functions, then

$ \int uv' dx = uv - \int u' v dx $

Examples

Integrate

$ \int x e^{3x} dx $

Solution

We use integration by parts. Notice that we need to use substitution to find the integral of e^x.

u = x	dv= e^3x dx
du = dx	v = 1/3 e^3x

Hence we have

$ \int x e^{3x} dx = x(\frac{1}{3} e^{3x}) - \int \frac{1}{3} e^{3x}dx $

$ = \frac{x e^{3x}}{3} - \frac{1}{9}e^{3x} + C $

Exercise

Evaluate

$ \int x ln x dx $

Integration By Parts Twice

Example

Evaluate

$ \int x^2 e^x dx $

We use integration by parts

u = x²	dv = e^x dx
du = 2x dx	v = e^x

We have

$ x^2 e^x - \int 2x e^x dx = x^2 e^x - 2 \int x e^x dx $

Have we gone nowhere? Now we now use integration by parts a second time to find this integral

u = x	dv = e^x dx
du = dx	v = e^x

We get

$ x^2 e^x - 2x e^x + 2\int e^x dx $

= x²e^x - 2xe^x + 2e^x + C

Other By Parts

Occasionally there is not an obvious pair of u and dv. This is where we get creative.

Example:

Find

$ \int ln x dx $

What should we let u and dv be? Try

u = ln x	dv = dx
du = 1/x dx	v = x

We get

$ x lnx - \int dx = x lnx - x + C $

When to Use Integration By Parts

When u-substitution does not work
When there is a mix of two types of functions such as an exponential and polynomial, polynomial and log, etc.
With ln x.
When all else fails.

Evaluating a Definite Integral

We can use integration by parts to evaluate definite integrals. We just have to remember that all terms receive the limits.

Example

Evaluate

$ \int_1^2 x^2 ln x dx $

Solution

Use integration by parts

u = ln x	dv = x² dx
du = 1/x dx	v = 1/3 x³

We get

$ \frac{1}{3}x^3 ln(x)|_1^e - \frac{1}{3}\int_1^e x^3 (\frac{1}{x}) dx = [\frac{1}{3}x^3 ln(x) - \frac{1}{9}x^3]_1^e $

$ = (\frac{1}{3} e^3 ln(e) - \frac{1}{9} e^3) - (\frac{1}{3} 1^3 ln(1) - \frac{1}{9} 1^3) = \frac{2}{9}e^3 + \frac{1}{9} $

Application: Present Value

Your patent brings you a annual income of 3,000 t dollars where t is the number of years since the the patent begins. The patent will expire in 20 years. A business has offered to purchase the patent from you. How much should you ask for it? Assume an inflation rate of 5%.

This question is a present value problem. Since there is inflation, your later earnings will be worth less than this year's earnings. The formula to determine this is given by

Present Value Formula

If c(t) is the continuous annual income over t₁ years with an inflation rate r, then the present value can by found by

$ \int_0^{t_1} c(t) e^{-rt} dt $

For our example, we have

c(t) = 2000 t r = 0.05 t₁ = 20

We integrate

$ \int_0^{20} 2000t e^{-0.05t} $

Use integration by parts and note that with the substitution

u = -0.05t du = -0.05dt

-20du = dt

we get

$ \int 2000 e^{-0.05t} = -20 \int e^u du $

$ = -20 e^u = -20 e^{-0.05t} $

so that

u = 2000t	dv = e^-0.05t dt
du = 2000 dt	v = -20e^-0.05t

This gives us

$ = [-40000t e^{-0.05t}]_0^{20 }+ 20000 \int_0^{20} e^{-0.05t} dt $

We have already found the antiderivative for this last integral. We have

$ = [-40000t e^{-0.05t}]_0^{20} - 800000 e^{-0.05t} ]_0^{20} = 211,393 $

You should ask for $211,393.

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\( \int uv' dx = uv - \int u' v dx \)

\( \int_0^{t_1} c(t) e^{-rt} dt \)