SRSG UNIVERSE-HPC Training

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:

$\displaystyle u=2x+1$

so $\displaystyle \frac{du}{dx}=2$ and $\displaystyle \frac{dx}{du}=1/2$

And on the second fraction:

$\displaystyle w=x-5$

so $\displaystyle \frac{dw}{dx}=1$ 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 $2\,$ cm $^2$ 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 $L$ as $\displaystyle k\,e^{-cL}$ , 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.

Integration 2

YouTube lecture recording from October 2020

Integration by Parts and by Partial Fractions

Integration reverses the process of differentiation

In general

Reminder: method 1: integration by substitution

Integration Method 2: Integration by Parts

Integration by parts: Example 1

Integration by parts: Example 2

Integration by parts: Example 3

Integration by parts: Example 4, trigonometry

Integration Method 3: Partial Fractions

Introductory problems

Introductory problems 1

Introductory problems 2

Introductory problems 3

Main problems

Main problems 1

Main problems 2

Main problems 2

Main problems 3

Extension problems

Extension problems 1

Extension problems 2

Extension problems 3