# Area under a curve

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• August 6th 2010, 12:38 PM
bobsanchez
Area under a curve
Find the area under the graph to the x axis from -2 to 2 for

f(x) = (4 - x^2)^(1/2)

Okay, so I understand how to do the problem overall but I don't know how to take the antiderivative of that particular function. Is there an easy way to do the chain rule in reverse or what...
• August 6th 2010, 12:44 PM
Ackbeet
Is that $f(x)=\sqrt{4-x^{2}}$? If so, I wouldn't integrate. You can solve the problem without integrating. Think about what the shape of that curve is.
• August 6th 2010, 12:53 PM
bobsanchez
Yeah, that's the function. It's a semicircle?
• August 6th 2010, 12:55 PM
Ackbeet
Quote:

It's a semicircle?
Show me that it's a semicircle.
• August 6th 2010, 01:08 PM
bobsanchez
It just is. What do you mean?
• August 6th 2010, 01:12 PM
Ackbeet
I asked my question because you seemed unsure of whether it's a semicircle or not. If you are sure, then what do you get for the area under it?
• August 6th 2010, 01:24 PM
bobsanchez
Oh, sorry. I had the question mark there because I didn't know where you were going with it, not whether or not it's a semicircle. Um...I don't really know. It's doesn't quite touch the x-axis according to my graphing calculator, and as a result it is sort of messing with all of the estimations I've attempted, and I don't know how to integrate a function like that.
• August 6th 2010, 01:25 PM
Ackbeet
Are you required to solve this problem using integration? Because I find that simply writing the answer down is a lot easier. What's the radius of this semicircle? And what's the area of a semicircle in general?
• August 6th 2010, 01:25 PM
11rdc11
Follow Ackbeet's advice. Integration is just the area under the curve so...
• August 6th 2010, 01:29 PM
bobsanchez
I think integration is preferred, but this method should suffice.
• August 6th 2010, 01:32 PM
Ackbeet
• August 6th 2010, 01:34 PM
bobsanchez
6.28
• August 6th 2010, 01:35 PM
Ackbeet
Or, more accurately, $2\pi.$
• August 6th 2010, 01:46 PM
bobsanchez
Touche, but it was supposed to be rounded to two decimal places, so I had to.
• August 6th 2010, 01:55 PM
Ackbeet
Then 6.28 might not be the more accurate answer, but it is the better answer. In engineering, at least, you better give an answer that your boss is happy with. If he wants decimals, give him decimals. Otherwise, it won't be as useful to him.
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