Video: EG19M1-DiffAndInt-Q14v3

EG19M1-DiffAndInt-Q14v3

02:09

Video Transcript

Evaluate the definite integral of sin of π‘₯ 𝑑π‘₯ from negative πœ‹ over two to πœ‹ over two.

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

So we need to evaluate at πœ‹ over two and negative πœ‹ over two. And then we will subtract them. So here we’ve evaluated at πœ‹ over two. And now we subtract, evaluating at negative πœ‹ over two. So notice we have to expand the subtraction sign. So this is really plus cos of negative πœ‹ over two and minus 𝑐, 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 we’ve seen it, the 𝑐s would simply cancel. So we really didn’t have to add in that plus 𝑐. But we might need to evaluate at negative cos πœ‹ over two and cos of negative πœ‹ over two. The cos of πœ‹ over two is zero. So we have negative zero, which really there’s no such thing. So we can leave the negative sign out. And then cos of negative πœ‹ 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, we’re looking at the area from negative πœ‹ over two to πœ‹ over two, so from here to here. This area would be negative because it’s underneath the π‘₯-axis. And this area would be positive because it’s above the π‘₯-axis. This is why our area actually cancels to get zero.

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