Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The Definite Integral - Problem 1

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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When a function is given by straight lines, you can use simple geometry to find the area underneath it. This is the same as finding the definite integral. Remember that the area of a rectangle A = wh, where w is the width and h is the height. The area of a triangle is A = ½bh, where b is the base and h is the height. Using these formulas, you can find the area underneath a curve without antidifferentiation.

Let's solve a problem that involves definite integrals. Recall that when f(x) is greater than or equal to 0, that is when it's non-negative on the interval a,b then the area under y equals f(x) is exactly equal to the definite integral. This is one way that we can use the definite integral. So let's solve a problem.

The graph of y equals g(x) consists of two line segments as shown to the right, so let's take a look. So this simple graph, this is my function y equals g(x). I'm asked to evaluate two integrals. First, the integral from 2 to 6 of g(x).

Now keeping in mind that it exactly equals the area under the curve, we can just use Geometry to evaluate these integrals. So let's start with the integral from 2 to 6. The interval that I'm dealing with here is this interval from 2 to 6. It's the area under the curve, so it's this area.

So if the integral exactly equals that area, I can just use Geometry it's the area of a triangle. The area of a triangle is 1/2 base times height. The base from 2 to 6 is 4. The height, the y coordinate of this point is 6, so the height is 6. So this is 1/2 of 4 times 6, 12. So that's the area of the triangle and therefore the value of the integral.

So sometimes you don't have to do any fancy calculus, you just need to know that the integral actually gives you the area under the curve. Now it does that only when the function g(x) is non-negative on that interval. But notice g(x) is non-negative from 2 to 6. If I had gone past 6, it would become negative, and this wouldn't work.

Let's take a look at another problem. The integral from -5 to 6 of g(x). If we look back at our picture, now we're taking the integral over this interval; from -5 all the way to 6. So I want the area under the curve from here all the way up to 6. So I want this area, as well as this area. You can see that because its just area. You can add the area of the rectangle, and the area of the triangle. So let me write that down equals area of rectangle plus area of triangle. We already found the area of the triangle to be 12, so I'll just write that down, +12.

What about the area of the rectangle? Let's take a look. So this rectangle's base is 7 from -5 to 2 is a distance of 7. It's height is 6, the y coordinate of this point is 6. So 6 times 7 is going to be 42. 6 times 7 first. So that's 42 plus 12, 54. So the value of this integral from -5 to 6 of g(x) is 54.

Just remember that value 54 is exactly the area under this curve, under y equals g(x) above the x axis from -5 to 6. So sometimes you can just use Geometry to find the value of the definite integral.

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