The fundamental theorem of calculus states that if a function f has an antiderivative F, then the definite integral of f from a to b is equal to F(b)-F(a).
Using this, you can solve for the area bounded by a curve and the x-axis. To do so, first find the points at which the graph crosses the x-axis by solving for x when f(x)=0. The results will be the limits of integration. Then, find the antiderivative, and take the integral of the function, and evaluate at the limits of integration solved for previously. Remember that when taking definite integrals, it is possible for the result to be negative (this means that the curve is below the x-axis).
Let’s solve one more problem using the fundamental theorem of calculus. Let’s use the fact that the derivative of xlnx minus x equals lnx to help me find the exact area under y equals lnx, this curve, from x equals 1 to x equals e.
Now the area of this region is going to exactly equal the integral from 1 to e of this function, lnx. All we really need is an antiderivative for lnx, but that’s what this formula gives me. Remember that every time you have a derivative formula, you also have an integral formula. This derivative formula means the integral of lnx is xlnx minus x, so this is an antiderivative for lnx. So I’m going to use that up here. The integral from 1 to e of lnxdx is xlnx minus x, evaluated from 1 to e.
First the e. This will be e times lne minus e. And then at 1, 1 times ln1 minus 1. Now lne is 1. E is the base of the natural log. And the log of its own base is 1. 1 times e is e, e minus e is 0. So this is zero in the first set of brackets. In the second set, Ln1 is 0, so all we have is -1 times -1. So minus -1, this area is exactly 1.
The area we were looking for, the area from 1 to e of under natural log x is exactly 1. Remember, we were able to solve this problem exactly and very easily using the fundamental theorem, just because we knew this one formula. So every time you have a new derivative formula, think about how it could be used as an integration formula, as an antiderivative to help you solve an area problem like this.