Definite integral of piecewise function | AP Calculus AB | Khan Academy

So we have an f of x right over here, and it's defined piecewise. For x less than zero, f of x is x plus one. For x greater than or equal to zero, f of x is cosine of pi x. We want to evaluate the definite integral from negative one to one of f of x dx.

You might immediately say, "Well, which of these versions of f of x am I going to take the anti-derivative from?" Because from negative 1 to 0, I would think about x plus 1, but then from 0 to 1, I would think about cosine pi x.

If you were thinking that, you're thinking in the right direction. The way that we can make this a little bit more straightforward is to actually split up this definite integral. This is going to be equal to the definite integral from negative 1 to 0 of f of x dx plus the integral from 0 to 1 of f of x dx.

Now why was it useful for me to split it up this way, in particular to split it up? I split the interval from negative one to one into two intervals: from negative one to zero and from zero to one. I did that because x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x.

So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. Then when you go from zero to one, f of x is cosine pi x. So, cosine of pi x.

Now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative 1 to 0 of x plus 1 dx.

Well, let's see, the anti-derivative of x plus 1 is... The anti-derivative of x is x squared over 2. I'm just incrementing the exponent and then dividing by that value, and then plus x. You could view this; I'm doing the same thing. If this is x to the 0, it'll be x to the first. x to the first over 1 is just x.

I'm going to evaluate that at 0 and subtract from that it evaluated at negative 1. And so this is going to be equal to... if I evaluate it at 0, let me do this in another color. If I evaluate it at 0, it's going to be 0 squared over 2, which is... well, I'll just write it: 0 squared over 2 plus 0. Well, all of that's just going to be equal to 0 minus it evaluated at negative 1.

So, minus negative 1 squared over 2 plus negative 1. So, negative 1 squared is just 1. So it's one-half plus negative 1. One-half plus negative one is... or one-half minus one is negative one-half. So all of that is negative one-half, but then we're subtracting negative one-half. Zero minus negative one-half is going to be equal to positive one-half.

So this is going to be equal to positive one-half. So this first part right over here is positive one-half, and now let's evaluate the integral from zero to 1 of cosine pi... I don't need that first parenthesis of cosine of pi x dx. What is this equal to?

Now, if we were just trying to find the anti-derivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x, but that's not what we have here. We have cosine of pi over pi x.

So there is a technique here; you could call it u substitution. You could say u is equal to pi x. If you don't know how to do that, you could still try to think about this where we could say, "All right, well maybe it involves sine of pi x somehow."

So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi.

Or you could say the derivative of sine pi x is pi cosine of pi x. Now we almost have that here, except we just need a pi. So what if we were to throw a pi right over here? But, so we don't change the value, we also multiply by 1 over pi.

So if you divide and multiply by the same number, you're not changing its value. 1 over pi times pi is just equal to 1. But this is useful because we now know that pi cosine pi x is the derivative of sine pi x.

So this is all going to be equal to... this is equal to 1. Let me take that 1 over pi. So this is equal to 1 over pi times... now we're going to evaluate. So the anti-derivative here we just said is sine of pi x, and we're going to evaluate that at one and at zero.

So this is going to be equal to 1 over pi times sine of pi... sine of pi minus sine of zero, which is just zero. Well, sine of pi, that's zero; sine of zero is zero. So you're going to have one over pi times zero minus zero, so this whole thing is just all going to be equal to zero.

So this first part was one-half; this second part right over here is equal to zero. So the whole definite integral is going to be one-half plus zero, which is equal to one-half.

So all of that together is equal to one-half.