In this section we learn to compute the value of a definite integral using the fundamental theorem of calculus.
1 The Definite Integral
We begin with the definition of the definite integral. Recall that a Riemann Sum for the function on the interval using sample points, , taken from intervals of width is given by For a non-negative function, a Riemann Sum gives area of rectangles. Due to the connection between these rectangles and the function , this sum approximates the area under the curve. Observing that the approximation approaches the exact value as the number of rectangles increases, we define the exact area under the curve as a limit of Riemann Sums and we call this limit the definite integral.
If the the graph of the function is a line or semi-circle, then we may be able to compute the definite integral by referring to a familiar geometric formula rather than referring to a Riemann Sum.
Definite Integrals that represent areas of familiar regions
For the next example, we need to know that for a number , the graph of the equation is a circle centered at the origin with radius . Solving this equation for , we get: The graph of is the upper semi-circle and the graph of is the lower semi-circle.
The above examples were very special in that the region under the curve had a familiar shape which made calculating the definite integral an exercise in using elementary geometry formulas. In general, the area cannot be computed in this elementary manner. The alternative is to use a limit of Riemann Sums. It turns out that this method is quite laborious (as we will show in the example below), but fortunately, there is a beautiful result which reduces the solution process to finding anti-derivatives. We will explore this Fundamental Theorem of Calculus in the next section.
This integral represents the area under the parabola over the interval . We will compute this definite integral using the Riemann Sums, as indicated in the definition of the definite integral. First, we subdivide the interval into equal sub-intervals, each of length whose endpoints have the form . Next, we choose the sample points to be right endpoints, so that . Then the Riemann Sum becomes We will evaluate this sum before letting . Using the formula our Riemann Sum becomes Finally, we can compute the definite integral by taking the limit as and noting that the polynomials in the formula above are both of degree 3: by simplifying the ratio of the lead coefficients, . As mentioned earlier, there is a more elegant method for solving this problem using the Fundamental Theorem of Calculus, presented in the next section.