Introduction To Integration

[Ted_Grendy]

Hello all

I was hoping someone could shed light on the following:-

Given the definite integral:-

Int (x^2+2) dx from [1] to [3]

This is asking me to find the area under the function (x^2+2) between 1 and 3.

Inorder to do this I need find the integral of the function which is:- (x^3)/3 + 2x

I then plug the values of 3 and 1 into (x^3)/3 + 2x and subtract.

Is this correct?

The reason why I am having trouble thinking it through is because I am trying to link why the area under a given function is the integral of the function which is a different function and it is the integral function to which we pass in the upper and lower limits - it seems bizarre.

Looking at this more deeply; then all functions can be considered as the integral of another function or the derivative of another function.

Or am I over thinking it?

Thoughts?

Thanks

 

[pka]

This is asking me to find the area under the function (x^2+2) between 1 and 3.
Inorder to do this I need find the integral of the function which is:- (x^3)/3 + 2x
I then plug the values of 3 and 1 into (x^3)/3 + 2x and subtract.
Is this correct? YES

\[\int_1^3 {\left( {{x^2} + 2} \right)dx = \left. {\frac{1}{3}{x^3} + 2x} \right|_{x = 1}^{x = 3} = \left[ {\frac{1}{3}{{(3)}^3} + 2(3)} \right] - \left[ {\frac{1}{3}{{(1)}^3} + 2(1)} \right] = \left[ {9 + 6} \right] - \left[ {\frac{1}{3} + 2} \right]}=~? \]
The above id standard for finding area. SO now ask your question, please.

 

[Dr.Peterson]

In order to do this I need find the integral of the function which is:- (x^3)/3 + 2x

I then plug the values of 3 and 1 into (x^3)/3 + 2x and subtract.

Is this correct? Yes.

The reason why I am having trouble thinking it through is because I am trying to link why the area under a given function is the integral of the function which is a different function and it is the integral function to which we pass in the upper and lower limits - it seems bizarre.

You're asking, I think, why the Fundamental Theorem of Calculus is true.

Briefly, think about what would happen if you moved your upper limit from 3 to something a little more than 3, say [MATH]3 + \Delta x[/MATH]. The area would increase by a little strip whose area is about [MATH]f(3)\Delta x[/MATH]. So the rate of change of area would be (approximately) [MATH]\frac{f(3)\Delta x}{\Delta x} = f(3)[/MATH]. So it sort of makes sense that the integrand, f, is the derivative of the area function, F. That is, the function you use to find the area, F(x), is the antiderivative (indefinite integral) of f(x).

For a fuller version of this argument, see Wikipedia's section on the geometric meaning of the FTC. Or, for more, read further down to see the proofs.

 

[pka]

The reason why I am having trouble thinking it through is because I am trying to link why the area under a given function is the integral of the function which is a different function and it is the integral function to which we pass in the upper and lower limits - it seems bizarre.
Looking at this more deeply; then all functions can be considered as the integral of another function or the derivative of another function.

Well, I had hoped to here from Grendy on this. Here is the historical basis.
Because of the way The Calculus is presented derivatives are taught before integrals. The historical facts are quite the opposite. Newton is given credit for giving us calculus in the seventeenth century. In his Method of Fluxions he considered a curve generated by a continuously moving point. As an astronomer he was interested in the area swept out by the movement of planets. He was able to establish the connection between the path and the speed. We now understand that to be the derivative/integral relation. The derivative of position is velocity, the derivative of velocity is acceleration. Ted, I don't know your "grade" level. But if you have use of a good mathematics library there are two books that can help: Calculus by Gillman (1973ed) and An Introduction To the History of Mathematics by Howard Eves.