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I'm covering some of my old calculus problems, and have one that I'm having trouble with, so I'm looking for a chat site. Anyone know of one?

Thanks.

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Last Post by hruzam
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Thanks, I figured someone here might want to take a shot at it, but I didn't want to post in the wrong place.

I'm not doing this for any class, I'm just reviewing my calc and physics to refresh my mind and get me back in the loop. The problem is fairly simple, but it's really giving me fits. It comes out of an older textbook, "Calculus, with Analytic Geometry", second edition by Earl W. Swokowski, page 250, exercise 7...I need to evaluate the definite integral from 1 to 2 of 5/(8)(x to the 6th power) dx. From my notes from years ago, I know the answer is 31/256, but I keep coming up with 29/256. Unfortunately, my notes have become fragmented over the years, and the work portion of my notes is gone. I'm also trying another problem...the definite integral from 4 to 9 of the function (t-3)/(the square root of t) dt. I don't show that as one of the problems I worked previously, so again I don't have my notes. I keep coming up with 55/18, when the answer should be 20/3. I believe my errors occur when raising the denominator portion of the function.

Anyone care to show me where I'm going wrong? I think once I get past these, I should be home free for another 100 pages or so.

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I need to evaluate the definite integral from 1 to 2 of 5/(8)(x to the 6th power) dx. From my notes from years ago, I know the answer is 31/256, but I keep coming up with 29/256.

Rewrite this as (5/8)*x^(-6) (where ^ is an operator indicating exponentiation). Then an antiderivative is (5/8)*(x^(-5)/(-5)) = -x^(-5) / 8. (I've omitted the 'plus C' that is usually written.) Then -(2)^(-5) / 8 - -(1)^(-5) / 8 = -1 / (32 * 8) + 1 / 8 = -1/256 + 32/256 = 31/256.

I'm also trying another problem...the definite integral from 4 to 9 of the function (t-3)/(the square root of t) dt. I don't show that as one of the problems I worked previously, so again I don't have my notes. I keep coming up with 55/18, when the answer should be 20/3. I believe my errors occur when raising the denominator portion of the function.

I'd begin by rewriting: (t - 3) / sqrt(t) = (t / sqrt(t)) - 3 / sqrt(t) = sqrt(t) - 3 / sqrt(t). Then I'd split this into two integrals.

The integral of t^(1/2) from 4 to 9 is 38/3, and the integral of 3 * t^(-1/2) is 6. Hence, the integral of t^(1/2) - 3 / t^(1/2) from 4 to 9 is 38/3 - 6, or 20/3.

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Rewrite this as (5/8)*x^(-6) (where ^ is an operator indicating exponentiation). Then an antiderivative is (5/8)*(x^(-5)/(-5)) = -x^(-5) / 8. (I've omitted the 'plus C' that is usually written.) Then -(2)^(-5) / 8 - -(1)^(-5) / 8 = -1 / (32 * 8) + 1 / 8 = -1/256 + 32/256 = 31/256.

I like your starting point but the next step was confusing.

Perhaps this might be more intuitive?

The other problem i do the same tho

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I did begin by rewriting as (5/8)*x^(-6). And actually, in one instance of working the antiderivative, I arrived at (5/8)*(x^(-5)/(-5). I made a stupid error in my math from that point, which resulted in a wrong answer. Now I've got it. I'm glad to see that my methodology was correct.

I haven't tried the second one yet, but in reviewing what you wrote, I believe I again made stupdi math errors which resulted in wrong answers. I think it just gets to the point where you see the situation so many times that you can no longer look at it objectively.

I appreciate the advise. I don't like proceding untiil I understand the present, so this will get me going again.

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Oh, and the constant, "+C" is not needed at this point. That's the next chapter on indefinite integrals.

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Yes, I had arrived at (-40) as the denominator also. It was easier to work by multiplying out the (8) and (-5). Thanks.

Now, how did you manage to post using proper math notation, such as the integral signs and ranges. Did you do this in another program, such as Word, and cut and paste? It does make the problem much easier to visualize.

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I'm guessing he used some plugin or something with Microsoft Word.

Another way is to use LaTeX, and he might have used that, but the font and size of the integral signs doesn't look right.

One easy way, without installing software, is to use Wikipedia. Click an edit button on this page and you can use Wikipedia's math notation feature -- you can preview your changes and save whatever images you see of your notation.

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I'm guessing he used some plugin or something with Microsoft Word.

Another way is to use LaTeX, and he might have used that, but the font and size of the integral signs doesn't look right.

One easy way, without installing software, is to use Wikipedia. Click an edit button on this page and you can use Wikipedia's math notation feature -- you can preview your changes and save whatever images you see of your notation.

Here's a more detailed example...

http://meta.wikimedia.org/wiki/Help:Formula#Integral_Equation

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Thanks, both of you. I'm sure I'll need some assistance in the future, and it's a lot easier using the proper notation.

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