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# Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area.$y = \sin x$, $0 \le x \le \pi$

## $$2$$

Integrals

Integration

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for this problem. We are examining the equation y equals sine of X, and we're looking at values of X from zero to pi inclusive. Now we want to find the area under this curve. We're going to do it two ways. First, we're going to actually look at the graph and kind of get an estimate of what we expect to see. Then we're going to use calculus to find the exact value. So first you can use a calculator graphing calculator, a graphic application on your computer, whatever you have that you're comfortable with. For this example, I am using the desk Most graphing application on my computer. I have programmed in Why equal sign effects, which is our equation from zero to pi, As you can see for decimus in order to give that range, I just put it inside curly brackets. Your application, if you use something different than the nomenclature, might be a little different. But you can see I have one hump going from the X axis back down to the X axis again. It's above the X axis. So I'm expecting a positive number. Well, if I look at this, if I imagine taking the piece The excess from 2 to 3, that little triangle piece that almost fits and makes a rectangle that would have a base of two and a height of one. I mean, it's not perfect, but it's awfully close. So I'm going to estimate Let me just write this down here. My estimate is that we're going to be around two units for our for our answer just from looking at that graph. Now, let's find our exact answer using calculus. So to find the area under the curve, we're going to take the integral. I'm evaluating this integral from X equaling zero to pi. So those are my limits of integration of sine x dx? Well, what is the integral of sine? The integral of sine is negative co sign of X okay. And I want to evaluate this from X equals zero two X equals pi. So let's put these in. We always start with our upper limit first. So if I sake excess pie co sign of pie is negative one. I'm taking the opposite of that. So that's a one. Now I subtract the value at the lower limit. X equaling zero co sign of zero is one. Take the opposite of that. It's going to be negative one. So one minus negative one. Well, that's adding. So actually, my answer is to my estimate, was exactly right.

Rochester Institute of Technology

Integrals

Integration

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