# 5.3 The fundamental theorem of calculus  (Page 2/11)

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$f\left(c\right)=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx,$

and the proof is complete.

## Finding the average value of a function

Find the average value of the function $f\left(x\right)=8-2x$ over the interval $\left[0,4\right]$ and find c such that $f\left(c\right)$ equals the average value of the function over $\left[0,4\right].$

The formula states the mean value of $f\left(x\right)$ is given by

$\frac{1}{4-0}{\int }_{0}^{4}\left(8-2x\right)dx.$

We can see in [link] that the function represents a straight line and forms a right triangle bounded by the x - and y -axes. The area of the triangle is $A=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right).$ We have

$A=\frac{1}{2}\left(4\right)\left(8\right)=16.$

The average value is found by multiplying the area by $1\text{/}\left(4-0\right).$ Thus, the average value of the function is

$\frac{1}{4}\left(16\right)=4.$

Set the average value equal to $f\left(c\right)$ and solve for c .

$\begin{array}{ccc}8-2c\hfill & =\hfill & 4\hfill \\ \hfill c& =\hfill & 2\hfill \end{array}$

At $c=2,f\left(2\right)=4.$

Find the average value of the function $f\left(x\right)=\frac{x}{2}$ over the interval $\left[0,6\right]$ and find c such that $f\left(c\right)$ equals the average value of the function over $\left[0,6\right].$

$\text{Average value}=1.5;c=3$

## Finding the point where a function takes on its average value

Given ${\int }_{0}^{3}{x}^{2}dx=9,$ find c such that $f\left(c\right)$ equals the average value of $f\left(x\right)={x}^{2}$ over $\left[0,3\right].$

We are looking for the value of c such that

$f\left(c\right)=\frac{1}{3-0}{\int }_{0}^{3}{x}^{2}dx=\frac{1}{3}\left(9\right)=3.$

Replacing $f\left(c\right)$ with c 2 , we have

$\begin{array}{ccc}{c}^{2}\hfill & =\hfill & 3\hfill \\ c\hfill & =\hfill & \text{±}\sqrt{3}.\hfill \end{array}$

Since $\text{−}\sqrt{3}$ is outside the interval, take only the positive value. Thus, $c=\sqrt{3}$ ( [link] ).

Given ${\int }_{0}^{3}\left(2{x}^{2}-1\right)dx=15,$ find c such that $f\left(c\right)$ equals the average value of $f\left(x\right)=2{x}^{2}-1$ over $\left[0,3\right].$

$c=\sqrt{3}$

## Fundamental theorem of calculus part 1: integrals and antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1 , is stated here. Part 1 establishes the relationship between differentiation and integration.

## Fundamental theorem of calculus, part 1

If $f\left(x\right)$ is continuous over an interval $\left[a,b\right],$ and the function $F\left(x\right)$ is defined by

$F\left(x\right)={\int }_{a}^{x}f\left(t\right)dt,$

then ${F}^{\prime }\text{(}x\right)=f\left(x\right)$ over $\left[a,b\right].$

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, $F\left(x\right),$ as the definite integral of another function, $f\left(t\right),$ from the point a to the point x . At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x , the definite integral is a number. So the function $F\left(x\right)$ returns a number (the value of the definite integral) for each value of x .

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.

find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.