Can someone help explain anti derivative and area?

[davidiswhat]

Can someone help explain anti derivative and area? Like in the notation i get why is f(x) used as basically the height of the curve and dx makes sense but why is the anti derivative the area? F of f' = f but why is that f the area? Also why are we making the function a derivative of another function?

 

[HallsofIvy]

Like in the notation i get why is f(x) used as basically the height of the curve and dx makes sense but why is the anti derivative the area? F of f' = f but why is that f the area? Also why are we making the function a derivative of another function?

I'm not sure you can get, here, any better explanation than that given in any Calculus text.

Given a function y= f(x), with f(x)> 0 for a< x< b, define \(\displaystyle F(x_0)\) to be "the area between the curve, y= f(x), and the x-axis between x= a and \(\displaystyle x= x_0\)". Now, imagine increasing \(\displaystyle x_0\) a little bit, to \(\displaystyle x_0+ h\). Then \(\displaystyle F(x_0+ h)\) is defined as "the area between the curve, y= f(x), and the x axis between x= a and \(\displaystyle x= x_0+ h\)". Then \(\displaystyle F(x_0+ h)- F(x)\) is "the area between the curve, y= f(x), and the x-axis between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\)".

Let p be the minimum value of F(x) between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\) and let q be the maximum value of f(x) between \(\displaystyle x_0\) and \(\displaystyle x_0+h\). That is, the graph of y= F(x) always lies above the horizontal line y= p and below the horizontal line y= q for x between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\). That further means that the area under the graph y= F(x), between \(\displaystyle x_0\) and \(\displaystyle x_0+h\) lies between the area of the rectangle with height q and base h (so area qh) and the area of the rectangle with height p and base h (so area qh): \(\displaystyle ph< F(x_0+h)- F(x_0)< qh\).

Dividing by h, \(\displaystyle p< \frac{F(x_0+h)- F(x_0)}{h}< q\).

Now take the limit as h goes to 0. Since p is the smallest value of f(x) in the interval \(\displaystyle x_0\) to \(\displaystyle x_0+ h\) and q is the largest value of f(x) in that interval, as h goes to 0, the interval reduces to the single point \(\displaystyle x_0\) and p and q both go to \(\displaystyle f(x_0)\). That is,
\(\displaystyle f(x_0)\le \lim_{h\to 0}\frac{F(x_0+h)- F(x_0)}{h}\le f(x_0)\)

Of course, that means that we must have \(\displaystyle \lim_{h\to 0}\frac{F(x_0+ h)- F(x_0)}{h}= f(x_0)\)

 

[davidiswhat]

I'm not sure you can get, here, any better explanation than that given in any Calculus text.

Given a function y= f(x), with f(x)> 0 for a< x< b, define \(\displaystyle F(x_0)\) to be "the area between the curve, y= f(x), and the x-axis between x= a and \(\displaystyle x= x_0\)". Now, imagine increasing \(\displaystyle x_0\) a little bit, to \(\displaystyle x_0+ h\). Then \(\displaystyle F(x_0+ h)\) is defined as "the area between the curve, y= f(x), and the x axis between x= a and \(\displaystyle x= x_0+ h\)". Then \(\displaystyle F(x_0+ h)- F(x)\) is "the area between the curve, y= f(x), and the x-axis between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\)".

Let p be the minimum value of F(x) between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\) and let q be the maximum value of f(x) between \(\displaystyle x_0\) and \(\displaystyle x_0+h\). That is, the graph of y= F(x) always lies above the horizontal line y= p and below the horizontal line y= q for x between \(\displaystyle x_0\) and \(\displaystyle x_0+ h\). That further means that the area under the graph y= F(x), between \(\displaystyle x_0\) and \(\displaystyle x_0+h\) lies between the area of the rectangle with height q and base h (so area qh) and the area of the rectangle with height p and base h (so area qh): \(\displaystyle ph< F(x_0+h)- F(x_0)< qh\).

Dividing by h, \(\displaystyle p< \frac{F(x_0+h)- F(x_0)}{h}< q\).

Now take the limit as h goes to 0. Since p is the smallest value of f(x) in the interval \(\displaystyle x_0\) to \(\displaystyle x_0+ h\) and q is the largest value of f(x) in that interval, as h goes to 0, the interval reduces to the single point \(\displaystyle x_0\) and p and q both go to \(\displaystyle f(x_0)\). That is,
\(\displaystyle f(x_0)\le \lim_{h\to 0}\frac{F(x_0+h)- F(x_0)}{h}\le f(x_0)\)

Of course, that means that we must have \(\displaystyle \lim_{h\to 0}\frac{F(x_0+ h)- F(x_0)}{h}= f(x_0)\)

That's a little too complex for me
I think i would get it better if you explained abstractly
Um let me simplify the question to why is ∫f(x)dx the anti derivative of f(x) (the area)?

 

[pka]

That's a little too complex for me
I think i would get it better if you explained abstractly
Um let me simplify the question to why is ∫f(x)dx the anti derivative of f(x) (the area)?

That's not surprising. Reply #2 is a correct explanation of the facts of the connection between the area bounded by a curve and the x-axis. If f(x) is a non-negative function on [a,b] and A(x) is the area from a to x under the function then A'(x)=f(x). A full blown development was worked out by Leonard Gillman in the 1960's(?) and is the basis of a calculus text book by Gillman & McDowell.

It is incorrect of you to think that "∫f(x)dx the anti derivative of f(x) (IS) (the area)".
Indefinite integrals are not about area but rather area is about definite integrals.
That is because the derivative of the area function is the function that determines the area.

Think of the graph of \(\displaystyle \sin(x)\) on \(\displaystyle [0,\pi]\).
The area from \(\displaystyle x=0\) to \(\displaystyle x=\frac{\pi}{2}\) is \(\displaystyle [-\cos(\pi/2)]-[-\cos(0)]\).
Finding the antiderivative is necessary for finding area, because the derivative of area is the defining function.