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[MUSIC]. 

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Let's use the fundamental theorem of 
calculus to evaluate an integral. 

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Let's do the integral from zero to pi of 
sine x dx. 

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Let's call this function little f. 
So, f of x, little f of x, will be sine 

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of x. 
Now, I need to find an anti-derivative 

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for little f, right? 
I need a function whose derivative is sin 

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of x. 
I know one already, right? 

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The derivative of minus cosine of x. 
Well, if the derivative of cosine is 

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minus sine, then I've got a minus sign 
there. 

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That's a little joke. 
All right? 

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The derivative of minus cosine of x is 
sine of x. 

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So, minus cosine of x is an 
anti-derivative for sine of x. 

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Now, what does the fundamental theorem of 
calculus tell us to do? 

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The fundamental theorem of calculus tells 
me that if I want to integrate this 

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function little f, I should find an 
antiderivative, which I've located, 

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right? 
Minus cosine, and then evaluate that 

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antiderivative at these end points and 
take the difference. 

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So, in this particular case what do I 
get? 

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So, in this case, I've got a function big 
F, right this is my antiderivative, and 

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it's minus cosine of x. 
And according to the fundamental theorem, 

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if I want to integrate from zero to pi, 
my function little f, which is sine x, 

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I'm going to evaluate the antiderivative 
at pi and subtract the antiderivative at 

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zero. 
In this case my chosen antiderivative is 

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minus cosine x. 
So, it's minus cosine pi minus, minus 

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cosine zero. 
Now, what's cosine of pi? 

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That's minus 1 but it's negative minus 1. 
That's 1 minus and cosine of zero is 1, 

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but it's negative 1. 
So, it's 1 minus negative 1, which is 2. 

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There are exactly 2 square units of area 
under the graph of sine of x. 

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between x equals 0 and x equals pi. 
can we actually see that fact? 

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Well, here's a graph of y equals sine x 
and by the interval that we just did, the 

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area under this curve is 2 square units. 
Well, here's 2 square units, right? 

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The maxium value of sine is 1. 
So, this is just at the same height as 

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where sine hits its maximum here. 
And its width, this rectangle is 2. 

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So, this is 2 square units. 
And the claim is that I should be able to 

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just about fit this under the graph. 
So, I've got my scissors here and I'll 

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just start cutting. 
And if I cut. 

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here, this part now, and it fits pretty 
well with the graph. 

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And I've got this little bit left over. 
Let me just. 

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Chop this off here. 
I mean this is, this is looking pretty 

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good. 
Right? 

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I mean that's not, it's not perfect but I 
mean, you know, it looks like about 2 

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square units fit underneath this graph. 
I always thought it was a little bit 

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surprising that the answer, the area 
under the graph of sine x, turned out to 

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be exactly 2 square units. 
I mean, you might have thought that the 

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answer would involve pi or something. 
And the answer ends up being a lot nicer 

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I think than you might have expected.