5.2 – The Definite Integral

Introduction Recall from the last section: Compute an area Try to find the distance traveled by an object

Definition Definition of a Definite Integral If f is a continuous function defined for a ≤ x ≤ b, we divide the interval [a,b] into n subintervals of equal width ∆x = (b – a)/n. We let x 0 ( = a), x 1, x 2, …, x n ( = b) be the endpoints of these subintervals and we let x 1 *, x 2 *, …, x n * be any sample points in these subintervals, so x i * lies in the ith subinterval [xi-1,xi]. Then the definite integral of f from a to b is Because f is continuous, the limit in this definition always exists and gives the same value no matter how we choose the sample points x i *.

Remarks We could replace x with any other letter without changing the value of the integral:

Riemann Sum Riemann sum: If f happens to be positive, then… - can be seen as the area under the curve y = f(x) from a to b:

Riemann Sum (cont’d) If f takes on both positive and negative values: The Riemann sum is the sum of the areas of The rectangles that lie above the x-axis and The negatives of the rectangles that lie below the x-axis; The limit of such Riemann sums gives the net area (the difference of the areas above and below the axis)

Remarks We have defined by dividing [a,b] into subintervals of equal width, but sometimes we would prefer to work with subintervals of unequal width. If the subinterval widths are ∆x 1, ∆x 2, …, ∆x n, we have to ensure that all these widths approach 0 in the limiting process.

Example 1 Express as an integral on the interval [0, π ].

Evaluating Integrals (DO NOT MEMORIZE)

Example 2(a) Evaluate the Riemann sum for f(x) = x 3 – 6x The sample points to be right endpoints and a = 0, b = 3, and n = 6

Example 2(b) Evaluate

Example 2 (solution)

Example 3 (a) Evaluate by interpreting it as an area.

Example 3(b) Evaluate by interpreting it as an area

The Midpoint Rule When the purpose is to find an approximation to an integral, it is usually best to choose the sample point x i * to be the midpoint of the interval, which we denote by x i. Any Riemann sum is an approximation to an integral, but we use midpoints we get the following approximation: Midpoint Rule Where And midpoint of

Example 4 Use Midpoint Rule with n = 5 to approximate