The fundamental theorem of calculus links the relationship between differentiation and integration. We have seen from finding the area that the definite integral of a function can be interpreted as the area under the graph of a function. It justifies our procedure of evaluating an antiderivative at the upper and lower bounds of integration and taking the difference.
The first part of the theorem shows that indefinite integration can be reversed by differentiation. It also proves that for every continuous function, there is an antiderivative (integral).
Let F be any antiderivative, or indefinite integral, for f on [a,b]. If f is a continuous function and is defined by
Every continuous function f has an antiderivative, and there is one antiderivative F(x) given by a definite integral of f with a variable upper boundary. Varying the lower boundary of the integrand will produce other antiderivatives.
One of the first things to notice about the fundamental theorem of calculus is that the variable of differentiation appears as the upper limit of integration in the integral. This makes sense because if we are taking the derivative of the integrand with respect to x, it needs to be in either (or both) the limits of integration. If we are integrating a function with respect to x, the function we end up with needs to be in terms of x.
The second part of this theorem tells us how to evaluate a definite integral assuming that f has an indefinite integral:
where F is any antiderivative (or indefinite integral) of continuous function f on a closed interval [a,b]. This part of the theorem is one that we use in finding area. It allows us to compute the definite integral of a function by using one of infinitely many antiderivatives, or indefinite integrals.
Here is a simple example of the fundamental theorem of calculus:
It is important to note that if F(x) is an antiderivative of f(x) and the function is defined on some interval, then every other antiderivative G(x) of f(x) differs from F(x) by a constant:
If we recall the general power rule for integration, taking the integral of a function will give us a constant. So there are infinitely many different antiderivatives for any given function.
Let's do a couple of examples using of the theorem.
Use the second part of the theorem and solve for the interval [a, x].
Taking the derivative with respect to x will leave out the constant.
Here is a harder example using the chain rule.
This is asking for the derivative of the integrand from the interval [2, sin(x)]. In other words, we need to find the original function.
From the first part of the fundamental theorem of calculus, we
Since sin(x) is in our interval, we let sin(x) take the place of x
We take the derivative of both sides with respect to x.
From the first part of the theorem, G'(x) = esin2(x) when sin(x) takes the place of x.
From chain rule, we take the derivative of the outside function (G) times the derivative of the inside function (sinx).
We then have our original function.