# Video: EG19M1-DiffAndInt-Q14v3

EG19M1-DiffAndInt-Q14v3

02:09

### Video Transcript

Evaluate the definite integral of sin of π₯ ππ₯ from negative π over two to π over two.

First, what is the antiderivative or the integral of sin of π₯? Itβs negative cos of π₯. Now normally with integrals, we have this plus π in our answer. And itβs okay to leave it in here like this. However, when we evaluate a definite integral, the πs end up cancelling. But we can leave it in here just to see.

So we need to evaluate at π over two and negative π over two. And then we will subtract them. So here weβve evaluated at π over two. And now we subtract, evaluating at negative π over two. So notice we have to expand the subtraction sign. So this is really plus cos of negative π over two and minus π, because if we subtract a negative, it becomes positive. Or we can imagine multiplying two negatives together make a positive. Imagining that, the subtraction sign was a negative one with multiplying.

So as weβve seen it, the πs would simply cancel. So we really didnβt have to add in that plus π. But we might need to evaluate at negative cos π over two and cos of negative π over two. The cos of π over two is zero. So we have negative zero, which really thereβs no such thing. So we can leave the negative sign out. And then cos of negative π over two is also zero. And zero plus zero gives us zero.

However, it may cross our minds that an integral is an area underneath a graph, underneath a curve. So how could we possibly get an area of zero? Well, if we were to plot the sine graph, it would resemble something like this. And remember, weβre looking at the area from negative π over two to π over two, so from here to here. This area would be negative because itβs underneath the π₯-axis. And this area would be positive because itβs above the π₯-axis. This is why our area actually cancels to get zero.