# 5.2 The definite integral  (Page 2/16)

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Previously, we discussed the fact that if $f\left(x\right)$ is continuous on $\left[a,b\right],$ then the limit $\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({x}_{i}^{*}\right)\text{Δ}x$ exists and is unique. This leads to the following theorem, which we state without proof.

## Continuous functions are integrable

If $f\left(x\right)$ is continuous on $\left[a,b\right],$ then f is integrable on $\left[a,b\right].$

Functions that are not continuous on $\left[a,b\right]$ may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.

It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

## Evaluating an integral using the definition

Use the definition of the definite integral to evaluate ${\int }_{0}^{2}{x}^{2}dx.$ Use a right-endpoint approximation to generate the Riemann sum.

We first want to set up a Riemann sum. Based on the limits of integration, we have $a=0$ and $b=2.$ For $i=0,1,2\text{,…,}\phantom{\rule{0.2em}{0ex}}n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of $\left[0,2\right].$ Then

$\text{Δ}x=\frac{b-a}{n}=\frac{2}{n}.$

Since we are using a right-endpoint approximation to generate Riemann sums, for each i , we need to calculate the function value at the right endpoint of the interval $\left[{x}_{i-1},{x}_{i}\right].$ The right endpoint of the interval is ${x}_{i},$ and since P is a regular partition,

${x}_{i}={x}_{0}+i\text{Δ}x=0+i\left[\frac{2}{n}\right]=\frac{2i}{n}.$

Thus, the function value at the right endpoint of the interval is

$f\left({x}_{i}\right)={x}_{i}^{2}={\left(\frac{2i}{n}\right)}^{2}=\frac{4{i}^{2}}{{n}^{2}}.$

Then the Riemann sum takes the form

$\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x=\sum _{i=1}^{n}\left(\frac{4{i}^{2}}{{n}^{2}}\right)\frac{2}{n}=\sum _{i=1}^{n}\frac{8{i}^{2}}{{n}^{3}}=\frac{8}{{n}^{3}}\sum _{i=1}^{n}{i}^{2}.$

Using the summation formula for $\sum _{i=1}^{n}{i}^{2},$ we have

$\begin{array}{cc}\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x\hfill & =\frac{8}{{n}^{3}}\sum _{i=1}^{n}{i}^{2}\hfill \\ \\ \\ \\ & =\frac{8}{{n}^{3}}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\hfill \\ & =\frac{8}{{n}^{3}}\left[\frac{2{n}^{3}+3{n}^{2}+n}{6}\right]\hfill \\ & =\frac{16{n}^{3}+24{n}^{2}+n}{6{n}^{3}}\hfill \\ & =\frac{8}{3}+\frac{4}{n}+\frac{1}{6{n}^{2}}.\hfill \end{array}$

Now, to calculate the definite integral, we need to take the limit as $n\to \infty .$ We get

$\begin{array}{cc}{\int }_{0}^{2}{x}^{2}dx\hfill & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x\hfill \\ \\ \\ & =\underset{n\to \infty }{\text{lim}}\left(\frac{8}{3}+\frac{4}{n}+\frac{1}{6{n}^{2}}\right)\hfill \\ & =\underset{n\to \infty }{\text{lim}}\left(\frac{8}{3}\right)+\underset{n\to \infty }{\text{lim}}\left(\frac{4}{n}\right)+\underset{n\to \infty }{\text{lim}}\left(\frac{1}{6{n}^{2}}\right)\hfill \\ & =\frac{8}{3}+0+0=\frac{8}{3}.\hfill \end{array}$

Use the definition of the definite integral to evaluate ${\int }_{0}^{3}\left(2x-1\right)dx.$ Use a right-endpoint approximation to generate the Riemann sum.

6

## Evaluating definite integrals

Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the x -axis.

## Using geometric formulas to calculate definite integrals

Use the formula for the area of a circle to evaluate ${\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}dx.$

The function describes a semicircle with radius 3. To find

${\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}dx,$

we want to find the area under the curve over the interval $\left[3,6\right].$ The formula for the area of a circle is $A=\pi {r}^{2}.$ The area of a semicircle is just one-half the area of a circle, or $A=\left(\frac{1}{2}\right)\pi {r}^{2}.$ The shaded area in [link] covers one-half of the semicircle, or $A=\left(\frac{1}{4}\right)\pi {r}^{2}.$ Thus,

$\begin{array}{}\\ \\ {\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}\hfill & =\frac{1}{4}\pi {\left(3\right)}^{2}\hfill \\ & =\frac{9}{4}\pi \hfill \\ & \approx 7.069.\hfill \end{array}$

why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg