Let be a continuous function on the interval . Divide the interval into subdivisions of equal width, , and choose a point in each interval. Then, the definite integral of from to is
We’ve been looking at two big things in this chapter: antiderivatives and the area under a curve. In the early days of the development of calculus, they were not known to be connected to one another. The integral sign wasn’t originally used in both instances. (Gottfried Leibniz introduced it as an elongated S to represent the sum when finding the area.) Connecting these two seemingly separate problems is done by the Fundamental Theorem of Calculus
Evaluate the following definite integrals. Include a sketch of the graph with the area you’ve found shaded in. Approximate the area to check to see if your definite integral answer makes sense. (Note: Just a guess, you don’t have to use Riemann sums. Use the grid to help.)
Explain how to compute the exact value of each of the following definite integrals using the Fundamental Theorem of Calculus. Leave all answers in exact form, with no decimal approximations.