definite integrals

Math

Definite integrals are the “area under the curve” part of calculus. It relates a bunch to anti-derivatives, which are the indefinite integrals.

The general idea is the definite integral is “area between F and zero when above the x-axis” - “area between F and zero when below the x-axis”.

It’s written like:

$$\begin{array}{r}{\int }_a^{b}f(x)dx\end{array}$$

A lot of the area functions, once we have definite bounds, begin to look a lot like geometry (which I don’t recall very well).

The properties of a definite integral

Let $f$ and $g$ be defined on a closed interval $I$ that contains the values $a$ and $b$ and $c$, and let $k$ be a constant. The following hold:

$$\begin{array}{rl}{\int }_a^{a}f(x)dx& =0\\ & \\ {\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx& ={\int }_a^{c}f(x)dx\\ & \\ {\int }_a^{b}f(x)dx& =-{\int }_b^{a}f(x)dx\\ & \\ {\int }_a^b(f(x)\pm g(x))dx& ={\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx\\ & \\ {\int }_a^{b}k\cdot f(x)dx& =k\cdot {\int }_a^{b}f(x)dx\end{array}$$

Solving w/ geometry

Circles

of note: circle w/ center (a,b) and radius r: $(x-a{)}^2+(y-b{)}^2=r^2$

Calculate: ${\int }_0^2\sqrt{4-x^2}dx$

$$\begin{array}{r}y=f(x)=\sqrt{4-x^2}\\ \\ y^2=4-x^2\\ \\ x^2+y^2=4\end{array}$$

That means we have a circle w/ radius 2 ($\sqrt{4}=2$) centered on 0,0 (b/c a=0, b=0).

To get the integral, we want the integral below the funciton, but above the x-axis, so it’s the northeastern quadrant.

So b/c that’s a quarter of a circle, $\frac{1}{4}\cdot \pi 2^2=\frac{4}{4}\pi =\pi$.

—

Calculate: ${\int }_1^{7}2+\sqrt{9-(x-4{)}^2}dx$

$$\begin{array}{r}y=f(x)=2+\sqrt{9-(x-4{)}^2}\\ \\ y-2=\sqrt{9-(x-4{)}^2}\\ \\ (y-2{)}^2=9-(x-4{)}^2\\ \\ (x-4{)}^2+(y-2{)}^2=9\end{array}$$

So the center of the circle is (4,2) with radius 3.

To get the integral, the origin is off to the side, so we don’t just take a quarter of the circle. The 2+ says we erase the parts under 2 from our graph. So it’s basically a half of the circle’s area, plus the rectangle from the 2 down to the x-axis. So $\frac{\pi \cdot 9}{2}+2\cdot 6$ because the 2*6 is the height under the curve, times the length of the curve. So the area is $12+\frac{9}{2}\pi$.

—

Calculate: ${\int }_3^{5}1+\sqrt{4-(x-3{)}^2}dx$.

$$\begin{array}{r}y=f(x)=1+\sqrt{4-(x-3{)}^2}\\ \\ y-1=\sqrt{4-(x-3{)}^2}\\ \\ (y-1{)}^2=4-(x-3{)}^2\\ \\ (x-3{)}^2+(y-1{)}^2=4\end{array}$$

This means that the origin is (3,1). Radius is 2.

So the origin of 3 mapping to the lower integral of 3 means we’re only concerned with the right half of the circle.

The 1+ means we’re only interested in things above 1, which makes it the top half of the circle.

$\text{area}=\frac{1}{4}(\pi \cdot 2^2)+1\cdot 2=2+\pi =5.14$

exercises in 5.2

5-13

5

given y= -2x+4

a) ${\int }_0^1(-2x+4)dx$

• it should be the triangle from 0-2 minus 1-2. So 0->2 would be $\frac{1}{2}\cdot 2\cdot 4-\frac{1}{2}\cdot 1\cdot 2$ = 4-1 = 3.

b) ${\int }_0^2$ = 4 c) $4+\frac{1}{2}\cdot 2\cdot -2$ = 2 (not 6.. right?) d) 1->3 1,2-> 1,-2 = 0 e) 2->4; 2*-4*.5 = -4 f) 0-1, but different formula (3x). 9.

7

y = f(x)

a) 0->2 = 2*4*.5 = 4 b) 2->4 = 2*2*.5 = 2 c) 2-4, 2f(x); 4 d) 0->1, 4x dx; 2 (b/c .5*1*4 = 2) e) 2->3; (2x-4)dx; 1*2*.5 = 1

When we can’t use geometry to find the answers

So when geometry isn’t easy (e.g. $y=x^2$) we can use Reimann Sums. This is “draw equally spaced bounding boxes on the curve, then get the area of the rectangles”. There are three common ways to place the boxes: left hand, right hand and midpoint. It’s a question of where you take the intersection point for a given interval.

If you have $[0,1]$ as the interval you’re calculating, you could take 0 (left-hand) as the representative value, you could take 1 (right-hand) or you could take 0.5 (midpoint).

To increase fidelity of the estimation provided by Reimann sums, you can add more rectangles. The syntax is weird, so we do summation notation which looks like ${\sum }_{i=1}^9{a}_i$.

To split the rectangles into equally spaced parts, we need some way of taking in $x$ and outputting some portion of it. The formula for grabbing that is:

$$x_i=x_1+(i-1)\Delta x$$

As an example, if we want to partition [0,4] into 100 intervals, we’d end up with $\Delta x=4/100=0.04$, so $x_{32}=x_1+31\cdot 0.04=1.24$.

The left hand rule is ${\sum }_{i=1}^{16}f(x_i)\Delta x$ Right hand: ${\sum }_{i=1}^{16}f(x_{i+1})\Delta x$ midpoint: ${\sum }_{i=1}^{16}f(\frac{x_i+x_{i+1}}{2})\Delta x$

Rules of summation

$$\begin{array}{rlr}\sum _{i=1}^{n}c& =c\cdot n& \text{(if c is constant)}\\ & & \\ \sum _{i=m}^n(a_i\pm b_i)& =\sum _{i=m}^n{a}_i\pm \sum _{i=m}^n{b}_i& \\ & & \\ \sum _{i=m}^{n}c\cdot a_i& =c\cdot \sum _{i=m}^n{a}_i& \\ & & \\ \sum _{i=m}^j{a}_i+\sum _{i=j+1}^n{a}_i& =\sum _{i=m}^n{a}_i& \\ & & \\ \sum _{i=i}^{n}i& =\frac{n(n+1)}{2}& \\ & & \\ \sum _{i=i}^n{i}^2& =\frac{n(n+1)(2n+1)}{6}& \\ & & \\ \sum _{i=i}^n{i}^3& =(\frac{n(n+1)}{2}{)}^2& \end{array}$$

Example of how to break down summation

$$\begin{array}{rl}\sum _{i=1}^6{a}_i& =\sum _{i=1}^6(2i-1)\\ & \\ & =2\cdot \sum _{i=1}^{6}i-\sum _{i=1}^{6}1\\ & \\ & =2\cdot (\frac{6(6+1)}{2})-1\cdot 6\\ & \\ & =2\cdot 21-6\\ & \\ & =36\end{array}$$

Example of approximating definite integrals using sums

example 5.3.4 from the book

$$\begin{array}{r}\sum _{i=1}^{16}f(x_{i+1})\Delta x\end{array}$$

so $\Delta x=\frac{4}{16}$ since the integral is 4 parts and we’re splitting it up into 16 chunks.

$$\begin{array}{rlr}{x}_i& =x_1+(i-1)\Delta x& \\ & & \\ & =0+(i-1)\Delta x& \text{because the start of the interval is 0}\end{array}$$

so $x_{i+1}=0+(i-1)+1\Delta x=i\Delta x$

So let’s break down ${\int }_0^4(4x-x^2)dx$ into those 16 chunks.2

$$\begin{array}{rl}{\int }_0^4(4x-x^2)dx& \approx \sum _{i=1}^{16}f(x_{i+1})\Delta x\\ & \\ & =\sum _{i=1}^{16}f(i\Delta x)\Delta x\\ & \\ & =\sum _{i=1}^{16}(4i\Delta x-(i\Delta x{)}^2)\Delta x\\ & \\ & =\sum _{i=1}^{16}(4i\Delta x^2-i^2\Delta x^3)\Delta x\\ & \\ & =(4\Delta x^2)\sum _{i=1}^{16}i-\Delta x^3\sum _{i=1}^{16}{i}^2\\ & \\ & =(4\Delta x^2)\frac{16\cdot 17}{2}-\Delta x^3\frac{16\cdot 17\cdot (2\cdot 16+1)}{6}\\ & \\ & =(4\cdot (\frac{4}{16}{)}^2)\cdot 136-(\frac{4}{16}{)}^3\cdot 1496\\ & \\ & =10.625\end{array}$$

Q: Why didn’t we just plug $\Delta x=0.25$ earlier? I think it’s because we need to figure out how to simplify $i$ first.

Reimann Sum definition

If $f$ is on a closed interval $[a,b]$ then $\Delta x$ is a partition of $[a,b]$ and $c_i$ denotes a value in the $i^{th}$ subinterval..

$$\begin{array}{r}\sum _{i=1}^{n}f(c_i)\Delta x_i\end{array}$$

is the Riemann Sum of $f$ on $[a,b]$.

Partitions of Reimann Sums

So not all partiitons need to be the same length.

The partition $\Delta x$ of a closed interval $[a,b]$ is $x_1,x_2,...x_{n+1}$ where $a=x_1<x_2<...<x_n<x_{n+1}=b$. The $i^{th}$ subinterval $[x_i,x_{i+1}]=x_{i+1}-x_i$.

If all of the partitions are equal length, $\Delta x$ represents the length of the intervals. $||\Delta x||$ represents the length of the largest subinterval of the partition.

When the intervals are equal length, their size is $\Delta x=\frac{b-a}{n}$. The $i^{th}$ term of the partition is $x_i=a+(i-1)\Delta x$

Reimann Sum practice

Algorithm

1. Find the size of the partitions
2. Find out what the interval is for a given partition $x_i$
3. Find the point (e.g. left hand rule vs midpoint, etc) in a given partition $x_i$.
4. Compute the reimann sum (remember, i=1 & n = number-of-partitions)
   1. plug in the “midpoint” equivalent formula for x
   2. simplify f via substitution
   3. simplify everything w/ rules of summation

example 5.3.5

Approximate ${\int }_{-2}^3(5x+2)dx$ using the midpoint rule and 10 equally spaced intervals.

I think we’re making

$$\begin{array}{r}\sum _{i=1}^{10}(5i+2)\cdot \Delta x\end{array}$$

So total size of a partition is: $\Delta x=\frac{3-(-2)}{10}=\frac{1}{2}$

The interval for a given partition is:

$$\begin{array}{rl}{x}_i& =a+(i-1)\Delta x\\ & \\ x_i& =-2+(i-1)\frac{1}{2}\\ & \\ x_i& =\frac{i}{2}-\frac{1}{2}\cdot -2+\frac{1}{2}\cdot -1\\ & \\ x_i& =\frac{i}{2}-\frac{5}{2}\end{array}$$

Midpoint:

​

\begin{align} point &= \frac{x_{i} + x_{i+1}}{2} \\ x_{i} &= \frac{i}{2} - \frac{5}{2} \\ x_{i+1} &= \frac{i+1}{2} - \frac{5}{2} = \frac{i}{2} - 2 \end{align}

I don’t really understand the jump for #3 just above. I know that the delta between i and i+1 = 1/2. Maybe (i+1)/2 breaks down into i/2+1/2 since it’s the same midpoint as before, but just one full interval forward. Then we basically take 1/2 from 5/2 which gives us 4/2 = 2.

Cool, so now we still need to find the midpoint.

$$\begin{array}{rl}& =\frac{\frac{i}{2}-\frac{5}{2}+\frac{i}{2}-2}{2}\\ & \\ & =\frac{\frac{i}{2}+-\frac{5}{2}+\frac{i}{2}+-2}{2}\\ & \\ & =\frac{\frac{i}{2}\cdot 2-\frac{9}{2}}{2}\\ & \\ & =\frac{i-\frac{9}{2}}{2}\\ & \\ & =\frac{i}{2}-\frac{9}{4}\end{array}$$

Now onto the actual sum!

$$\begin{array}{rl}{\int }_{-2}^3(5x+2)dx& \approx \sum _{i=1}^{10}f(\frac{x_i+x_{i+1}}{2})\Delta x\\ & \\ & =\sum _{i=1}^{10}f(\frac{i}{2}-\frac{9}{4})\Delta x\\ & \\ & =\sum _{i=1}^{10}(5(\frac{i}{2}-\frac{9}{4})+2)\Delta x\\ & \\ & =\Delta x\sum _{i=1}^{10}(5\cdot (\frac{i}{2}-\frac{9}{4})+2)\\ & \\ & =\Delta x\sum _{i=1}^{10}(\frac{5}{2}\cdot i-\frac{37}{4})\\ & \\ & =\Delta x\cdot (\frac{5}{2}\sum _{i=1}^{10}i-\sum _{i=1}^{10}\frac{37}{4})\\ & \\ & =\Delta x\cdot (\frac{5}{2}\cdot \frac{10\cdot 11}{2}-\frac{370}{4})\\ & \\ & =\frac{1}{2}\cdot (\frac{5}{2}\cdot \frac{10\cdot 11}{2}-\frac{370}{4})\\ & \\ & =22.5\end{array}$$

5.3.6

Approximate ${\int }_0^4(4x-x^2)dx$

We know from 5.3.1 that:

$$\begin{array}{rl}\Delta x& =\frac{4-0}{n}=\frac{4}{n}\\ & \\ x_i& =0+\Delta x\cdot (i-1)=\frac{4(i-1)}{n}\\ & \\ \text{Right hand rule uses }{x}_{i+1}& =\frac{4i}{n}\end{array}$$ $$\begin{array}{rl}{\int }_0^4(4x-x^2)dx& \approx \sum _{i=0}^n\frac{4i}{n}\Delta x\\ & \\ & =\sum _{i=0}^n(4(\frac{4i}{n})-(\frac{4i}{n}{)}^2)\Delta x\\ & \\ & =\Delta x\cdot (\sum _{i=0}^n[(\frac{16}{n}\cdot i)-(\frac{16}{n^2}\cdot i^2)])\\ & \\ & =\Delta x\cdot (\frac{16}{n}\cdot \sum _{i=0}^{n}i-\frac{16}{n^2}\sum _{i=0}^n{i}^2)\\ & \\ & =\Delta x\cdot (\frac{16}{n}\cdot \frac{n\cdot (n+1)}{2}-\frac{16}{n^2}\cdot \frac{n(n+1)(2n+1)}{6})\\ & \\ & =\frac{4}{n}\cdot (\frac{16}{n}\cdot \frac{n\cdot (n+1)}{2}-\frac{16}{n^2}\cdot \frac{n(n+1)(2n+1)}{6})\\ & \\ & =\frac{64}{n^2}\cdot \frac{4n\cdot (n+1)}{2n}-\frac{64}{n^3}\cdot \frac{4n(n+1)(2n+1)}{6n}\end{array}$$ $$\begin{array}{rl}& =\frac{16n\cdot (n+1)}{2n}-\frac{16n(n+1)(2n+1)}{6n^2}\\ & \\ & =\frac{16n^2+16n}{2n}-\frac{(16n^2+16n)(2n+1)}{6n^2}\\ & \\ & =\frac{16(n+1)}{2}-\frac{32n^3+16n^2+32n^2+16n}{6n^2}\\ & \\ & =8(n+1)-\frac{32n^3+48n^2+16n}{6\cdot n\cdot n}\\ & \\ & =8(n+1)-\frac{16n^2+24n+8}{3\cdot n}\\ & \\ & =8(n+1)-\frac{16}{3}\cdot n+\frac{24}{3}+\frac{8}{n}\end{array}$$

urg. Honestly a bit lost.

Limits of Reimann Sums

$$\begin{array}{rl}\underset{n->\infty }{\lim }{S}_L(n)& =\underset{n->\infty }{\lim }{S}_R(n)=\underset{n->\infty }{\lim }{S}_M(n)=\underset{n->\infty }{\lim }\sum _{i=1}^{n}f(c_i)\Delta x\\ & \\ \underset{n->\infty }{\lim }\sum _{i=1}^{n}f(c_i)\Delta x& ={\int }_a^{b}f(x)dx\\ & \\ \underset{||\Delta x||->0}{\lim }\sum _{i=1}^{n}f(c_i)\Delta x_i& ={\int }_a^{b}f(x)dx\end{array}$$

Approximating curves using trapezoids

Each side of the trapezoids are used twice, except the first and last. So we can simplify our manual summations by this formula:

$$\begin{array}{rl}{\int }_a^{b}f(x)dx& \approx \sum _{i=1}^n\frac{f(x_i)+f(x_{i+1})}{2}\Delta x\\ & \\ & =\frac{\Delta x}{2}\sum _{i=1}^n(f(x_i)+f(x_{i+1}))\\ & \\ & =\frac{\Delta x}{2}[f(x_1)+2\cdot \sum _{i=1}^n(f(x_i))+f(x_{n+1})]\end{array}$$

So, given that $\Delta x=\frac{3\pi }{40}\approx 0.236$

${\int }_{\frac{-\pi }{4}}^{\frac{\pi }{2}}sin(x^3)dx\approx \frac{0.236}{2}[-0.466+2\cdot (-0.165+(-0.031)+...)+(-0.67)]=0.4275$

^^ For that to really work, we should build up a table of values so we know how to sum them.

Simpson’s rule

The idea is that using rectangles is a constant function, using trapezoids is a linear function and using simpson’s rule is using quadratic functions.

I don’t quite understand how they got this formula, despite them saying “it’s not hard to show that..”

​ $$\begin{array}{r}{\int }_{x_1}^{x_3}f(x)dx=\frac{x_3-x_1}{6}(y_1+4y_2+y_3)\end{array}$$

Because we need 3 points, that means it uses 2 partitions to do it, resulting in $n/2$ parabolas.

When we get the area under the curve, we use the formula:

$\frac{\Delta x}{3}[f(x_1)+4f(x_2)+2(x_3)+4f(x_4)+2(x_5)+\text{..repeat 4 then 2..}+f(x_{n+1})]$

Example 5.5.5

Approximate ${\int }_0^1{e}^{-x^2}$ into 4 subintervals.

Make a table of values:

i	$x_i$	$e^{-x_i^2}$
1	0	1
2	0.25	.9394
3	0.5	.7788
4	0.75	.570
5	1.0	.3678

$$\begin{array}{r}{\int }_0^1{e}^{-x^2}dx\approx \frac{0.25}{6}\cdot (1+4\cdot (.9394)+2\cdot (.7788)+4\cdot (.570)+.3678)=0.7469\end{array}$$

5.5.6

approximate ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}sin(x^3)dx$ into 10 equally spaced intervals using simpson’s rule

formula for $\Delta x=\frac{b-a}{n}$ $\Delta x=\frac{\frac{\pi }{2}+\frac{\pi }{4}}{n}\approx 0.236$

Table of values: (can’t seem to calculate sin(x^3) w/ formulas: https://emacs.stackexchange.com/questions/71332/sin-function-in-org-calc-being-weird)

i	$x_i$	$sin(x^3)$
0	-0.785	-0.46509031
1	-0.549	-0.16471509
2	-0.313	-0.030659492
3	-0.077	-4.5653298e-4
4	0.159	4.0196682e-3
5	0.395	0.061590868
6	0.631	0.24860482
7	0.867	0.60655029
8	1.103	0.97392189
9	1.339	0.67493118
10	1.575	-0.69281953

$$\begin{array}{r}\frac{0.236}{3}\cdot (-.466+4\cdot -.165+2\cdot .031+0+2\cdot .004+4\cdot .061+2\cdot .246\\ \\ +4\cdot .601+2\cdot .971+2\cdot .690+-.670)\\ \\ =0.4701\end{array}$$

Error bounds

Okay, so our approximations may be bad. To figure out how bad, we can use error bounds for the trepezoidal rule or simpson’s rule:

1. Let $E_T$ be the error in approximating ${\int }_a^{b}f(x)dx$ using the trapezoid rule w/ $n$ subintervals. If $f$ has a continuous 2nd derivative on [a,b] and M is any upperbound of $|f^{\prime \prime }(x)|$ on [a,b], then

$$\begin{array}{r}{E}_T\le \frac{(b-a{)}^3}{12n^2}M\end{array}$$

1. Let $E_S$ be the error in approximating ${\int }_a^{b}f(x)dx$ using Simpson’s rule w/ $n$ subintervals. If $f$ has a continuous 4th derivative on [a,b] and M is any upperbound of $|f^{(4)}|$ on [a,b] then

$$\begin{array}{r}{E}_S\le \frac{(b-a{)}^5}{180n^4}M\end{array}$$

Quadratic Formula

$$\begin{array}{r}\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$