So far, we know that calculating the integral (or antiderivative) of a function gives a general function that represents the area under the curve of that function. What if we want to determine the exact area under a function on a given interval? We use the Fundamental Theorem of Calculus (FTC) to do this!
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Jetzt kostenlos anmeldenSo far, we know that calculating the integral (or antiderivative) of a function gives a general function that represents the area under the curve of that function. What if we want to determine the exact area under a function on a given interval? We use the Fundamental Theorem of Calculus (FTC) to do this!
The FTC enables us to solve slightly more complex problems that are otherwise possible. For example, it's easy to determine who will travel farther in 30 minutes if you're going 10 miles per hour and your friend is moving at 12 miles per hour. But what if your velocities are not constant and are modeled by an equation that changes with respect to time? How would you answer this? Luckily, we have the FTC to allow us to do this. It can be split up into 2 parts. Let's see how it works!
If the function f is continuous on [a, b], then the function g, defined by
where ,
is continuous on [a, b] and differentiable on (a, b), and .
Put into words, the derivative of a definite integral with respect to the upper limit equals the integrand evaluated at the upper limit. Integration and differentiation are inverses!
The derivative of the integral of the function In layman's terms in equation form:
The derivative of the integral of a function results in the function itself! So, all part 1 of the FTC is saying is that integration and differentiation are inverses of each other - they undo each other!
This part is informally referred to as "the evaluating part of the theorem".
If the function f is continuous on [a, b], then
where F is the antiderivative of f, or a function such that .
Essentially, Part 2 is the inverse of Part 1. If the antiderivative F of f is already known, we can evaluate the integral by subtracting the values of F at the endpoints of the interval of integration.
First, we need to start off with a few pre-requirements. We need:
Remember, we have to prove that the derivative of the integral on the function f is the function f itself. In other words
Let x and be in the interval (a, b). Then, by the definition of the derivative, we have
Factoring out h leaves us with
For this next part, we need to use our Mean Value Theorem for Integrals muscles, found here! The area under any two points of a curve, let's say between points a and b, is calculated as
Rewriting the equation above In math notation according to the diagram below, we have
Rewriting the formula above, subbing in x for a, for b, and h for , we have
What happens as ? Two things happen:
Mathematically:
Ok, let's review what we have determined so far. First, we know that
Subbing equation II into equation I we get;
But wait, equation 1 above tells us what is! It's ! So
From the start of this proof, we saw that
Therefore, the derivative of is
Finally, we have
Let . Part 1 of FTC says that
If F is the antiderivative of f on the interval , then
where C is a constant.
Let's say we want to find the area under the curve between the same point. This area should be zero since there is no width! So the area at this point, let's say , is calculated as
Using the fact that , subbing in and , and using the fact that gives us
Calculus is an extremely powerful tool for evaluating integrals; it allows us to evaluate integrals without approximations or geometry. The main application of the FTC is finding exact integral answers. However, using the FTC, we can also find and study antiderivatives more abstractly.
As previously mentioned, the FTC helps us to solve more complex problems. Believe it or not, integration and the Fundamental Theorem of Calculus can help us calculate the center of mass, which is essentially a balancing point. If you stand and try to balance on one leg, you have to adjust your center of mass to balance on one leg rather than two. In Calculus, the center of mass of a graphical region is the point where the region is perfectly horizontally balanced. The FTC can help us to solve for the center of mass.
Let's first do 3 examples using FTC part 1!
Let's start with a simple example.
Find the derivative of and evaluate .
is continuous on the interval . Plugging in any number between 1 and infinity will not produce any discontinuities.
By the Fundamental Theorem of Calculus, all we need to do is replace the t variable of the inner function with x such that
Now that we have g'(x), we simply plug in 10 to find g'(10).
Now that we better understand the Fundamental Theorem of Calculus, let's move on to a more complex example involving The Chain Rule. If you haven't gone over the chain rule, please check out our article before proceeding!
Find the derivative of .
We know that the cosine function has no discontinuities.
We let , then
By the Fundamental Theorem of Calculus, all we need to do is replace the r variable of the inner function with u(x) such that
We must also apply the Chain Rule, so
Let's try an example with two variable limits of integration.
Evaluate
There is no such number that will produce a discontinuity when plugged in.
Since both limits of integration are variables, we must split the integral in two and use our definite integral rules to match the form of the Fundamental Theorem of Calculus.
We let , then
By the Fundamental Theorem of Calculus, all we need to do is replace the x variable of the inner function with t and u(x) such that
Again, we must also apply the Chain Rule, so
Now let's try an example using part 2 of the FTC!
Use the Fundamental Theorem of Calculus Part 2 to evaluate
Since the function inside the integrand is a polynomial, we know that it is continuous over the entire interval.
The antiderivative of the function is
To apply part 2 of the FTC, we simply plug in and into the antiderivative and subtract.
The Fundamental Theorem of Calculus shows the relationship between the integral and the derivative. Essentially, the theorem states that the derivative of a definite integral with respect to the upper limit is the same as the integrand evaluated at the upper limit.
You can use the formula for part 2 of the theorem to calculate the area under the curve between 2 points.
The function f must be continuous over the interval of integration.
Taking the derivative of an integral gives the original function that was being integrated.
The derivative of the integral from 0 to x of the function f(t)=t is x.
Why is the Fundamental Theorem of Calculus so important?
What is the Fundamental Theorem of Calculus in words?
Part 1 defines the relationship between the derivative and the integral. If you take the integral of a derivative f'(x), you will get the original function f.
Part two says you can find the area under a function using a formula
What is the requirement for the Fundamental Theorem of Calculus to be applied?
The function f must be continuous over the closed interval of integration
Given an antiderivative F(t) = t + 4, what is the area under the curve on the interval [0, 5]?
Using the second part of the FTC:
F(5) - F(0) = (5 + 4) - (0 + 4)
= 9 - 4
= 5
The Mean Value Theorem for integrals is derived from which two theorems?
State the Mean Value Theorem for integrals in words.
The area under a function curve is equal to the area of a rectangle whose width is the interval of the function and whose height is the average value of the function over the interval.
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