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Integration 2

YouTube lecture recording from October 2020

Integration by Parts and by Partial Fractions

Integration reverses the process of differentiation

Integration Method 2: Integration by Parts

Integration Method 3: Partial Fractions

Integration 2

Learning outcomes

Be able to use integration by parts to integrate a product of functions
Be able to apply the method of partial fractions to integrate fractional functions

YouTube lecture recording from October 2020

The following YouTube video was recorded for the 2020 iteration of the course. The material is still very similar:

Integration by Parts and by Partial Fractions

Integration reverses the process of differentiation

In general

$\displaystyle {d(a\thinspace x^{n+1})\over dx}=(n+1)a\thinspace~x^n~~~~~\Rightarrow ~~~~\int_{x_1}^{x_2} a\thinspace x^n~dx = \biggr[{a\over(n+1)} \thinspace x^{(n+1)}\biggl]_{x_1}^{x_2}$

Reminder: method 1: integration by substitution

This method can be thought of as an integral version of the chain rule. Suppose we wish to integrate:

$\displaystyle I=\int f(g(x))~dx= \int f(u)~dx$ > $\displaystyle I =\int f(u){dx\over du}~du$

Integration Method 2: Integration by Parts

Recall that:

$\displaystyle {d\over dx}f(x)~g(x)=f(x)~g'(x)+ g(x)~f'(x)$

Integrate and rearrange to get:

$\displaystyle \int\limits_a^b f(x)~g'(x)~dx=\biggr[\;f(x)~g(x)\;\biggl]_a^b-\int\limits_a^b g(x)~f'(x)~dx$

This is known as the formula for 'integrating by parts' and can also be written:

$\displaystyle \int\limits_a^b u~{dv\over dx}~dx ~~= ~~ \biggr[uv\biggl]^b_a-\int\limits_a^b v~{du\over dx}~dx$

$\displaystyle \int\limits_a^b u~v'~dx~~ =~~ \biggr[uv\biggl]^b_a-\int\limits_a^b v~u'~dx$

with

f(x) \equiv u

and

g(x) \equiv v

Integration by parts: Example 1

$\displaystyle \int\limits_a^b x~\sqrt{(x+1)}~dx=\int\limits_a^b x~(x+1)^{1/2}~dx$

Choose

$\displaystyle u=x\qquad{\rm and}\qquad v'=\sqrt{(x+1)}$

so that:

$\displaystyle u'=1\qquad{\rm and}\qquad v={2 \over 3}(x+1)^{3/2}$ .

Choose

$\displaystyle u=x\qquad{\rm and}\qquad v'=\sqrt{(x+1)}$

so that:

$\displaystyle u'=1\qquad{\rm and}\qquad v={2 \over 3}(x+1)^{3/2}$

Then:

$\displaystyle \int\limits_a^b x~\sqrt{(x+1)}~dx =\biggr[x~{2\over 3}(x+1)^{3/2}\biggl]_a^b - \int\limits_a^b 1 \times {2\over 3}(x+1)^{3/2}~dx$ > $\displaystyle =\biggr[{2\over 3}~x~(x+1)^{3/2}\biggl]_a^b- \biggr[{2\over 3}\cdot {2\over 5}(x+1)^{5/2}\biggl]_a^b$

If we had chosen the other option for

u

and

v'

we would have got:

$\displaystyle \biggr[(x+1)^{1/2}~{x^2\over 2}\biggl]_a^b - \int\limits_a^b {1\over 2}~{1\over \sqrt{(x+1)}}~{x^2\over 2}~dx$

The second term is worse than the integral we started with! It's important to choose

u

and

v

carefully. This comes with practice.

Integration by parts: Example 2

Let us calculate the indefinite integral,

$\displaystyle \int \ln x~dx$

We can do a bit of a trick. Let:

$\displaystyle u=\ln x \quad\Longrightarrow\quad u'={1\over x}\qquad{\rm and}\qquad v'=1\Longrightarrow v=x$

Putting these into our equation:

$\displaystyle \int \ln x~dx\quad=\quad\ln x~\cdot x - \int x~\frac{1}{x}dx$ > $\displaystyle = \quad x \ln x - \int 1 dx\quad=\quad x \ln x - x + C$

We can also check this calculation using SymPy:

import sympy as sp
x = sp.symbols('x')
sp.integrate(sp.log(x),x)

$\displaystyle x \log{\left(x \right)} - x$

Integration by parts: Example 3

$\displaystyle \int\limits^{\infty}_0 x^3~e^{-x}~dx$

Using the formula:

$\displaystyle \biggl(\int u\thinspace v'\thinspace dx = [u\thinspace v]-\int v\thinspace u'\thinspace dx\biggr)$

Choose

$\displaystyle u=x^3\qquad{\rm and}\qquad v'=e^{-x}$

so that:

$\displaystyle u'=3x^2\qquad{\rm and}\qquad v=-e^{-x}$

Then:

$\displaystyle \int\limits^{\infty}_0 x^3~e^{-x}~dx=-\biggr[x^3~e^{-x}\biggl]^{\infty}_0 +\int\limits^{\infty}_0 3x^2~e^{-x}~dx$

Now apply integration by parts to the right-hand side:

Choose

$\displaystyle u=3x^2\qquad{\rm and}\qquad v'=e^{-x}$

so that:

$\displaystyle u'=6x\qquad{\rm and}\qquad v=-e^{-x}$

Then:

$\displaystyle \int\limits^{\infty}_0 3x^2~e^{-x}~dx=-\biggr[3x^2~e^{-x}\biggl]^{\infty}_0 +\int\limits^{\infty}_0 6x~e^{-x}~dx$

And, once more:

Choose

$\displaystyle u=6x\qquad{\rm and}\qquad v'=e^{-x}$

so that:

$\displaystyle u'=6\qquad{\rm and}\qquad v=-e^{-x}$

Then:

$\displaystyle \int\limits^{\infty}_0 6x~e^{-x}~dx =-\biggr[6x~e^{-x}\biggl]^{\infty}_0 +\int\limits^{\infty}_0 6~e^{-x}~dx$ > $\displaystyle \Longrightarrow \int\limits^{\infty}_0 6~e^{-x}~dx=-\biggr[6~e^{-x}\biggl]^{\infty}_0=-6e^{-\infty}+6e^0=6$

(Since

e^{-\infty}=0

and

e^0=1

)

The other terms all go to zero:

$\displaystyle -\bigr[x^3~e^{-x}\bigl]^{\infty}_0 =-{\infty}^3~e^{-\infty} + 0 =0$ > $\displaystyle -\bigr[3x^2~e^{-x}\bigl]^{\infty}_0 =-3{\infty}^2~e^{-\infty} + 0 =0$

So, to answer our original question:

$\displaystyle \int\limits^{\infty}_0 x^3~e^{-x}~dx=6$

Let's also check it with SymPy:

sp.integrate(x**3 * sp.exp(-x),(x,0,sp.oo))

$\displaystyle 6$

This result actually generalises:

$\displaystyle \int\limits^{\infty}_0 x^n~e^{-x}~dx=n!$

Integration by parts: Example 4, trigonometry

Recall that:

$\displaystyle {d\over dx}(\sin x)=\cos x$ > $\displaystyle {d\over dx}(\cos x)=-\sin x$

Let's try and calculate the following integral:

$\displaystyle \int\limits^b_a~\cos x\;e^{-x}~dx = I$

Choose

$\displaystyle u=\cos x\qquad{\rm and}\qquad v'=e^{-x}$

so that:

$\displaystyle u'=-\sin x\qquad{\rm and}\qquad v=-e^{-x}$

Then:

$\displaystyle I =\biggr[-\cos x~\; e^{-x}\biggl]^b_a -\int\limits^b_a ~(-)\sin x~(-)e^{-x}~dx$ > $\displaystyle I =\biggr[-\cos x~\; e^{-x}\biggl]^b_a~~-~~\int\limits^b_a ~(-)\sin x~(-)e^{-x}~dx$

Next, choose

$\displaystyle u=\sin x\qquad{\rm and}\qquad v'=e^{-x}$

so that:

$\displaystyle u'=\cos x\qquad{\rm and}\qquad v=-e^{-x}$

Then:

$\displaystyle I =\biggr[-\cos x~\;~e^{-x}\biggl]^b_a~~-~~\biggr[\sin x~(-)e^{-x}\biggl]^b_a ~~+~~\int\limits^b_a ~\cos x~(-)e^{-x}~dx$

The last term is the integral we started with:

$\displaystyle \Longrightarrow~~~2~\int\limits^b_a ~\cos x~e^{-x}~dx~ =~\biggr[\sin x~\; e^{-x}\biggl]^b_a~~ -~~\biggr[\cos x~\; e^{-x}\biggl]^b_a$

Integration Method 3: Partial Fractions

If we want to calculate the integral below, none of the previous rules allow us to make much progress.

$\displaystyle \int {dx \over (2x+1)(x-5)}$

But, in this case, we can try splitting the denominator up into two fractions that we can deal with:

$\displaystyle {\rm Let~~~~~} {1 \over (2x+1)(x-5)}={A\over (2x+1)} + {B\over (x-5)}$

If we multiply both sides by

(2x+1)(x-5)

, we get:

$\displaystyle A(x-5)+B(2x+1)=1$ so $Ax-5A+B2x+B=1$

We can then equate coefficients of

x

$\displaystyle A+2B=0~~~~{\rm thus~~} A=-2B~~~~~~~~~~\rm (A)$

Equate units (coefficients of

x^0

$\displaystyle -5A+B=1 ~~{\rm~~so~from~(A):~~} 10B+B=1,~~B={1\over 11}~,~~A=-{2\over 11}$

Thus:

$\displaystyle \int {dx \over (2x+1)(x-5)}=-\int {2 dx\over 11(2x+1)} + \int {dx\over 11(x-5)}$

Now use method of substitution on the first fraction:

$\displaystyle u=2x+1$

\displaystyle \frac{du}{dx}=2

and

\displaystyle \frac{dx}{du}=1/2

And on the second fraction:

$\displaystyle w=x-5$

\displaystyle \frac{dw}{dx}=1

and

\displaystyle \frac{dx}{dw}=1

$\displaystyle \int {2 du \over 2 \times 11 \times u} + \int {dw\over 11w}= -{\ln u\over 11} + {\ln w \over 11}$ > $\displaystyle =-{\ln |2x+1|\over 11} + {\ln |x-5| \over 11}$

SymPy can also solve integrals requiring partial fractions:

sp.integrate(1/((2*x + 1)*(x - 5)),x)

$\displaystyle \frac{\log{\left(x - 5 \right)}}{11} - \frac{\log{\left(x + \frac{1}{2} \right)}}{11}$

This answer seems different because of the arbitrary constant of integration.

Introductory problems

Introductory problems 1

By using suitable substitutions, evaluate the following integrals:

$\displaystyle \def\d#1{{\rm d}#1} \int x^2(x^3+4)^2~~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int e^{-x}(5-4e^{-x})~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int (1+x)\sqrt{(4x^2+8x+3)}~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int 3x e^{(x^2+1)}~\d{x}$

Introductory problems 2

Find the indefinite integrals, with respect to

x

, of the following functions:

$\displaystyle x\,e^{3bx}$
$\displaystyle x^3\,e^{-3x}$
$\displaystyle x \cos (x)$
$\displaystyle e^{bx} \sin(x)$

Introductory problems 3

Sketch the curve

y=(x-2)(x-5)

and calculate by integration the area under the curve bounded by

x=2

and

x=5

Main problems

Main problems 1

Evaluate the following indefinite and definite integrals:

$\displaystyle \def\d#1{{\rm d}#1} \int \frac{6}{(7-x)^3}~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int 13x^3(9-x^4)^5~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int_2^5 5\log(x)~\d{x}$
$\displaystyle \def\d#1{{\rm d}#1} \int x^x\,(1 + \log(x))~\d{x}$

Main problems 2

Suppose the area

A(t)

(in cm

^2

) of a healing wound changes at a rate

$\displaystyle \def\dd#1#2{{\frac{{\rm d}#1}{{\rm d}#2}}} \dd{A}{t} = -4t^{-3},$

where

t

, measured in days, lies between 1 and 10, and the area is

2\,

^2

after 1 day. What will the area of the wound be after 10 days?

Main problems 2

A rocket burns fuel, so its mass decreases over time.

If it burns fuel at a constant rate

\rho\,{\rm kg/s}

, and if the exhaust velocity relative to the rocket is a constant

v_e\,{\rm m/s}

, then there will be a constant force of magnitude

\rho v_e

propelling it.

The rocket starts burning fuel at

t=0\,{\rm s}

with total mass of

m_0\,{\rm kg}

, and runs out of fuel at a later time

t=t_f\,{\rm s}

, with a final mass of

m_f\,{\rm kg}

Newton's second law tells us that the instantaneous acceleration $a$ of the rocket at time $t$ is equal to the force propelling it at that time, divided by its mass at that time. Write down an expression for $a$ as a function of $t$ .
By integrating this expression, show that the rocket's total change in velocity is given by $\displaystyle v_e \ln\left({m_0\over m_f}\right).$

Main problems 3

The flow of water pumped upwards through the xylem of a tree,

F

, is given by:

$\displaystyle F = M_0(p+qt)^{3/4},$

where

t

is the tree's age in days,

p

and

q

are positive constants, and

M_0p^{3/4}

is the mass of the tree when planted (i.e.\ at

t=0

Determine the total volume of water pumped up the tree in its tenth year (ignoring leap years) if:

$p=10$ ,
$q=0.01\,$ day $^{-1}$ , and
$M_0=0.92\,$ l $\,$ day $^{-1}$ .

Extension problems

Extension problems 1

Express

\displaystyle \frac{1}{x(x^2-16)}\quad{\rm in~the~form}\quad\frac{A}{x} + \frac{B}{(x+4)} + \frac{C}{(x-4)}

Hence calculate

\displaystyle \def\d#1{{\rm d}#1} \int\frac{1}{x(x^2-16)}\,\d{x}.

Extension problems 2

The probability that a molecule of mass

m

in a gas at temperature

T

has speed

v

is given by the Maxwell-Boltzmann distribution:

$\displaystyle f(v) = 4 \pi \left({m\over 2\pi k T} \right)^{3/2} v^2 e^{-mv^2/2kT}$

where

k

is Boltzmann's constant. Find the average speed:

$\displaystyle \def\d#1{{\rm d}#1} \overline {v} =\int_0^{\infty}v\,f(v)\,\d{v}.$

Extension problems 3

Baranov developed expressions for commercial yields of fish in terms of lengths,

L

, of the fish. His formula gave the total number of fish of length

L

\displaystyle k\,e^{-cL}

, where

c

and

k

are constants (

k

is positive).

Give a sketch of the graph $\displaystyle f(L)=k\,e^{-cL}$ . (Something decreasing, concave upward and asymptotic to horizontal axis will do.) On your sketch, introduce marks on the horizontal axis that represent lengths $L=1, L=2, L=3, L=4 {\rm and } L=5$ . Now draw a rectangle on your sketch that represents the number of fish whose lengths are between $L=3$ and $L=4$ .
Explain how we can represent the total number of fish $N$ as an area. Show that this number equals $k/c$ .
Only fish longer than $L_0$ count as commercial. Hence, assuming that the fish are all similar in shape (i.e. their width and breadth scales with their length) and of equal density $\rho$ , show that the weight, $W$ , of the commercial fish population is

$\displaystyle \def\d#1{{\rm d}#1} W= \int_{L_0}^{+\infty} a\, k \rho\,L^3 e^{-cL}\,\d{L},$

and hence that

$\displaystyle W={N\, a\, \rho\, e^{-cL_0}\over c^3} \left((cL_0)^3 +3(cL_0)^2+ 6cL_0 +6\right),$

where $a$ is a constant.