## Tuesday, December 12, 2006

### 5.3 The Fundamental Theorem of Calculus

Hey guys this is Izzy. Today in class we learned about the fundamental theorem of calculus. This theorem establishes a connection between the two branches of calculus: differential calculus and integral calculus. In class we covered how if you multiply the function that you are taking the integral of by a certain factor, that you can factor out that number. Example:

This makes it easier to deal with the original funtion first, without any coefficients to clutter up the process.

In class, we saw how the integral of a function can be a function itself, represented by g(x). Expressing the integral of a function as the function g(x) allows one to actually graph and express the area between the curve of a graph and the x-axis in terms of x:

In this graph,

By using this equation, we can find the area between the graph and the x-axis. Let's say we wanted to find the area between x=0 and x=3. You would multiply the change in x, 3, by f(3), 3, and .5, because it's a triangle. So, g(3)=(3)(3)(.5)=4.5 .

Today we also made the amazing discovery that the antiderivative of a function is the same as the integral of the function. Observe:

As you can see from the graph, the area of the shaded region (teal) is:

We established that

To find g'(x), we have to replace t with x and multiply whatever you get by the derivative of x.

Wait a minute, the derivative of the Area function is also x! Thus A'=g'(x). We just proved that the antiderivative of a function is the same as its integral! i'm so happy.

What we just did relates to the Fundamental theorem of Calculus, Part 1, which states that:
If the function f is continuous on [a,b], then the function g defined by

is continuous on [a,b] and differentiable on (a,b), and g'(x)=f(x)

MOVING ON, we now know how to find the derivative of these types of functions. Let's try this problem: Find

Whenever you see these types of problems, you must first look at the upper bound, or any of the bounds that has a variable in it. In the process of calculating the equation for g'(x), you must replace the variable t with the variable bound, and wherever you see dt, replace it with the derivative of the variable bound. In this case, the upper bound has the variable: x^2.
To solve this problem, we have to use the chain rule (make u=x^2):

There you have it.

This a reminder to Ami to do the next blog!

These are a few websites that help out with this concept: