[MUSIC].
Let's use the fundamental theorem of calculus to evaluate an integral.
Let's do the integral from zero to pi of sine x dx.
Let's call this function little f. So, f of x, little f of x, will be sine
of x. Now, I need to find an anti-derivative
for little f, right? I need a function whose derivative is sin
of x. I know one already, right?
The derivative of minus cosine of x. Well, if the derivative of cosine is
minus sine, then I've got a minus sign there.
That's a little joke. All right?
The derivative of minus cosine of x is sine of x.
So, minus cosine of x is an anti-derivative for sine of x.
Now, what does the fundamental theorem of calculus tell us to do?
The fundamental theorem of calculus tells me that if I want to integrate this
function little f, I should find an antiderivative, which I've located,
right? Minus cosine, and then evaluate that
antiderivative at these end points and take the difference.
So, in this particular case what do I get?
So, in this case, I've got a function big F, right this is my antiderivative, and
it's minus cosine of x. And according to the fundamental theorem,
if I want to integrate from zero to pi, my function little f, which is sine x,
I'm going to evaluate the antiderivative at pi and subtract the antiderivative at
zero. In this case my chosen antiderivative is
minus cosine x. So, it's minus cosine pi minus, minus
cosine zero. Now, what's cosine of pi?
That's minus 1 but it's negative minus 1. That's 1 minus and cosine of zero is 1,
but it's negative 1. So, it's 1 minus negative 1, which is 2.
There are exactly 2 square units of area under the graph of sine of x.
between x equals 0 and x equals pi. can we actually see that fact?
Well, here's a graph of y equals sine x and by the interval that we just did, the
area under this curve is 2 square units. Well, here's 2 square units, right?
The maxium value of sine is 1. So, this is just at the same height as
where sine hits its maximum here. And its width, this rectangle is 2.
So, this is 2 square units. And the claim is that I should be able to
just about fit this under the graph. So, I've got my scissors here and I'll
just start cutting. And if I cut.
here, this part now, and it fits pretty well with the graph.
And I've got this little bit left over. Let me just.
Chop this off here. I mean this is, this is looking pretty
good. Right?
I mean that's not, it's not perfect but I mean, you know, it looks like about 2
square units fit underneath this graph. I always thought it was a little bit
surprising that the answer, the area under the graph of sine x, turned out to
be exactly 2 square units. I mean, you might have thought that the
answer would involve pi or something. And the answer ends up being a lot nicer
I think than you might have expected.