Class 12 math (India)
- The fundamental theorem of calculus and definite integrals
- Intuition for second part of fundamental theorem of calculus
- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Definite integrals: reverse power rule
- Definite integral of rational function
- Definite integral of radical function
- Definite integral of trig function
- Definite integral involving natural log
- Definite integrals: common functions
- Area using definite integrals
Sal finds the definite integral of 9sin(x) between 11π/2 and 6π.
- [Voiceover] Let's see if we can evaluate the definite integral from 11 pi over two to six pi of nine sine of x dx. So the first thing, let's see if we can take the antiderivative of nine sine of x, and we could use some of our integration properties to simplify this a little bit. So this is going to be equal to, this is the same thing as nine times the integral from 11 pi over two to six pi of sine of x dx. And what's the anitderivative of sine of x? Well we know, from our derivatives, that the derivative with respect to x of cosine of x is equal to negative sine of x, negative sine of x. So can we construct this in some way so this is a negative sine of x? What if I multiplied, on the inside, what if I multiplied it by negative one? Well I can't just multiply in only one place by negative one, I need to multiply by negative one twice so I'm not changing its value. So what if I said negative nine times negative sine of x? Well this is still gonna be nine sine of x. If you took negative nine times negative sine of x, it is nine sine of x, and I did it this way because now negative sine of x, it matches the derivative of cosine of x. So we could say that this is all going to be equal to, it's all going to be equal to, you have your negative nine out front, negative nine times, times, I'll put it in brackets, negative nine times, the antiderivative of negative sine of x, well that is just going to be cosine of x, cosine of x, and we're going to evaluate it at its bounds. We're going to evaluate it at six pi, let me do that in a color I haven't used yet, we're gonna do that at six pi, and we're also going to do that at 11 pi over two, 11 pi over two, so this is going to be equal to, this is equal to negative nine times, I'm going to create some space here, so actually that's probably more space than I need, it's going to be cosine of six pi, cosine of six pi, minus cosine of 11 pi over two, cosine of 11 pi over two. Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could view six pi as, we're going around the unit circle three times. So this is the same thing as cosine of two pi, or the same thing as cosine of zero, so that is going to be equal to one. If that seems unfamiliar to you I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some multiple of two pi here to put it in values that we can understand better. So this is, so let me write it here, cosine of 11 pi over 2, that is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus, this is the same thing as five and one half pi, right? Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five, no, let's subtract four pi, which is eight pi over 2. So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine, and so this is going to be equal to cosine of 3 pi over 2. And if we imagine the unit circle, and let me draw the unit circle here. So it's my y axis, my x axis, and then I have the unit circle, so, whoops. So the unit circle just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go to three pi over two, so that's this point on the unit circle, so the cosine is the x coordinate, so this is going to be zero. This is zero so this is zero, and so we get one minus zero, so everything in the brackets evaluates out to one, and so we are left with, so let me do that, so all of this is equal to one. And so you have negative nine times one, which of course is just negative nine, is what this definite integral evaluates to.