Study Guide

# Definite Integrals - Definite Integrals of Real-Valued Functions

## Definite Integrals of Real-Valued Functions

When we're integrating a non-negative function from a to b, the integral can be thought of as the "area under the curve" of the function. However, most of the time we can't count on having a non-negative function to integrate.

Assume f is a function that's allowed to take on negative values, and we're integrating from a to b with a < b. Then is the weighted sum of the areas between the graph of f and the x-axis. We look at all areas between f and the x-axis. If they're on top of the x-axis we count them positively. If they're below the x-axis we count them negatively.

In other words, we add all the areas on top of the x-axis, then subtract all the areas below the x-axis.

• ### Conditions for Integration

We can only integrate real-valued functions that are reasonably well-behaved. No Dance Moms allowed. If we want to take the integral of f(x) on [a, b], there can't be any point in [a,b] where f zooms off to infinity. When it comes to definite integrals, this is bad: Depending on how the function behaves near the asymptote, we may still be able to take the integral, but it won't be a definite integral. These are called improper integrals, but we won't go in depth with those guys just yet.

Having a continuous function is great, but if it's only discontinuous at a few points, that's allowed too. For example, what if f(x) = 5 for all x ≠ 1 but is undefined at x = 1? That function looks like this: If we want to integrate f from 0 to 2 there's one little spot where f isn't defined. That means the integral needs to account for all the area in this rectangle except for the line at x = 2: Since a line doesn't have any area, taking out that line doesn't take away any area from the rectangle. This means it's not a problem for f to be undefined at that one point.

Similarly, it's not a problem for f to be undefined at ten separate points. Each individual line has no area, so the ten lines together have no area. It's also not a problem for f to be undefined at 100 separate points. Or a million.

When we can find the integral of a function on [a, b], we say that function is integrable on [a, b]. If a function is integrable for any interval we pick we say that function is integrable.

• ### General Riemann Sums

For those functions whose integrals we can't find exactly, we can still use the left-hand sum, right-hand sum, midpoint sum and trapezoid sum to estimate their integrals.

The left-hand, right-hand, and midpoint sums are examples of Riemann Sums.

A Riemann Sum is any sum you get when you split up [a, b] into sub-intervals. The intervals don't necessarily all have to be the same size. Draw a rectangle on the sub-interval using a value of the function on that sub-interval for the rectangle "height."

We put "height" in quotes because now the functions are allowed to take on negative values. While we can't have negative heights, we can think of a function value as a weighted height. If the weighted height is negative and the sub-interval goes from left to right, then the weighted area will also be negative.

Whether the LHS, RHS, MID, and TRAP are over- or under-estimates doesn't depend at all on whether f is positive or negative.

### Sample Problem

Let f be negative and increasing. The right-hand sum rectangles don't cover enough area, but when f is negative this means the right-hand sum will be less negative than the actual integral. Therefore the right-hand sum will be an over-estimate.

In order to determine if your sum gives an over-or under-estimate, you can rely on the things we learned earlier:

• If f is increasing then the left-hand sum is an under-estimate and the right-hand sum is an over-estimate.
• If f is decreasing then the left-hand sum is an over-estimate and the right-hand sum is an under-estimate.
• If f is concave up then the trapezoid sum is an over-estimate and the midpoint sum is an under-estimate.
• If f is concave down then the trapezoid sum is an under-estimate and the midpoint sum is an over-estimate.

It doesn't matter for any of these if f is positive or negative. Also, remember that you don't need to remember all of this. You can get by with this list and knowing how to draw graphs.

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