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[edit] February 14

[edit] Three calculus problems (studying for a test)

1. I need to evaluate the indefinite integral ∫x2√(1-x)dx. I tried using u-substitution:
u = 1 - x
x = 1 - u
du = -dx
dx = -du


∫(1 - u)2u1/2•-du + C
-∫(1 - 2u + u2)u1/2du + C
-(∫u1/2du - ∫2u3/2du + ∫u5/2du) + C
-∫u1/2du + ∫2u3/2du - ∫u5/2du + C
-2u3/2/3 + 4u5/2/5 - 2u7/2/7 + C
-2(1 - x)3/2/3 + 4(1 - x)5/2/5 - 2(1 - x)7/2/7 + C


The textbook says the answer is -2/105(1 - x)3/2(15x2 + 12x + 8) + C. Is this just an equivalent form that I'm not recognizing?


2. This time it's a definite integral: ∫19 1/[(√x)(1 + √x)2]dx. Again I tried u-substitution:
u = 1 + √x
du = (1/2)x-1/2dx = [1/(2√x)]dx


2∫u(1)u(9) u-2du
2∫24 u-2du
2•(u-3/-3)]24
2[4-3/-3 - 2-3/-3] -(2/3)(1/64 - 1/8)


But the book claims that the answer is 1/2! (That exclamation mark denotes my frustration, not a factorial.)


3. I need to find the value of c guaranteed by the mean-value theorem for integrals for the function f(x) = 2sec2 x over the interval [-π/4, π/4].
f(c) = [1/(b - a)]∫abf(x)dx
f(c) = [1/(π/4 - [-π/4])]∫-π/4π/4(2sec2) x)dx
f(c) = [1/(π/2)]2 • (tan x)]-π/4π/4
f(c) = (4/π)[tan (π/4) - tan (-π/4)]
f(c) = (4/π)[1 - (-1)]
f(c) = 8/π


Yet the official answer given is ±arccos (√π)/2 ≈ ±0.4817 ... what am I doing wrong??


Any help is much appreciated. Thanks so much. --anon


Hello. Yes, the first is an equivalent form - just factor out the (1-x)1.5 and it should be obvious. In the second, the substitution is fine, but you integrated u-2 to give u-3/-3, when the correct integral is -u-1 since you add one to the power, then divide through by it. As for question 3, your derivation is correct, but you missed the final step. You say f(c) = 8/pi, but you're asked to give c, not f(c).
Hope this helps. -mattbuck (Talk) 01:29, 14 February 2008 (UTC)
Dude, I love you. --anon —Preceding unsigned comment added by 141.155.31.240 (talk) 02:24, 14 February 2008 (UTC)

[edit] Surd Troubles

I've tried and i thought i did it correctly but as it seems i did not =[. Some help would be appreciated.

  • Problem 1 \sqrt{63}-\sqrt{28}=\sqrt{x}
    =\sqrt{9}\times\sqrt{7}-\sqrt{4}\times\sqrt{7}
    =3\sqrt{7}-2\sqrt{7}
    =1\sqrt{7}
    =\sqrt{7}
    but the textbook says =5\sqrt{7}
  • Problem 2 - Simplify 4\sqrt{7}\times(-2\sqrt{7})

=4\sqrt\times(-2\sqrt)=(-8\sqrt)
=\sqrt{7}\times\sqrt{7}=\sqrt{49}
=-8\sqrt{49}

  • Problem 3 - Simplify 2\sqrt{3}(4\sqrt{3}-\sqrt{6})

=8\sqrt{9}-\sqrt{18}
=8\times 3\times 3 - 9 \times 3
=24\sqrt{3}- 3\sqrt{3}
=21\sqrt{3}
textbook says =24 - 6\sqrt{2}
Thanks for the help =] !. —Preceding unsigned comment added by 121.219.227.217 (talk) 07:52, 14 February 2008 (UTC)

In problem 1, it's probably supposed to be a plus sign. In problem 2, since 49 is a square you can simplify it further. In problem 3, I think you should just do it more carefully. Black Carrot (talk) 08:02, 14 February 2008 (UTC)

Ah i figured out problem 2 and possibly 1( before i noticed your comment that'll teach me to refresh eh) [Note:In problem 1 a minus sign is written in the text book]
  • Problem 2 - Simplify 4\sqrt{7}\times(-2\sqrt{7})

=\sqrt{7}\times\sqrt{7}=\sqrt{7}^2=7
=4\sqrt\times(-2\sqrt)=(-8\sqrt)
=-8\times 7
= ( − 56) (this is the answer in the text book)

  • Problem 1 \sqrt{63}-\sqrt{28}=\sqrt{x}


=\sqrt{9}\times\sqrt{7}-\sqrt{4}\times\sqrt{7}
=9\sqrt{7}-4\sqrt{7} ( i know it isnt simplifying it but its all i could think of)
=5\sqrt{7}

And i got futher on problem three but still hit a snag..
  • Problem 3 - Simplify 2\sqrt{3}(4\sqrt{3}-\sqrt{6})

=8\sqrt{9}-\sqrt{18}
=\sqrt8\times \sqrt3\times \sqrt3 - \sqrt9 \times \sqrt2
=24 - 3\sqrt{2}
(Why isnt this 6!) textbook says =24 - 6\sqrt{2}
—Preceding unsigned comment added by 121.219.227.217 (talk) 08:25, 14 February 2008 (UTC)

For problem 1 - just because the textbook made a mistake, doesn't mean you have to, too. \sqrt{63}-\sqrt{28} is equal to \sqrt{7}. If the textbook says something else then it is wrong, and you shouldn't invent new math rules to cover up for it.
For problem 3 - take a closer look at your very first step. -- Meni Rosenfeld (talk) 08:33, 14 February 2008 (UTC)
Ah i see the problem i didn't know you had to do that, well i did learn something new now.

=8\sqrt{9}-2\sqrt{18}
=\sqrt8\times \sqrt3\times \sqrt3 -\sqrt2 \times \sqrt9 \times \sqrt2
=24 - 6\sqrt{2}

Thanks Guys =] you've been a big help! —Preceding unsigned comment added by 121.219.227.217 (talk) 08:49, 14 February 2008 (UTC)

Of the three expressions above, the first and last are equal, but the middle one is different. There are too many root signs there.  --Lambiam 18:37, 14 February 2008 (UTC)
Something else seems wrong, problem 1 says: \sqrt{63}-\sqrt{28}=\sqrt{x}, do they want you to simplify the left side or solve for x on the right (they're very different problems). If the book answer is to be believed then I suspect they asked you to simplify \sqrt{63}-\sqrt{28}. A math-wiki (talk) 02:17, 15 February 2008 (UTC)
I suspect this was indeed the question, but \sqrt{63}-\sqrt{28} is not equal to 5\sqrt{7}. -- Meni Rosenfeld (talk) 09:21, 15 February 2008 (UTC)
You seem to be having trouble formatting your equations. Do you need any help? Black Carrot (talk) 07:57, 15 February 2008 (UTC) 07:48, 15 February 2008 (UTC)

[edit] relation between arithmatic mean, median and mode

what is the relation between arithmatic mean,median and mode? what are its limitations?please explain in detail how the relation is derivedKasiraoj (talk) 14:18, 14 February 2008 (UTC)

Mode (statistics), Arithmetic mean, Median explains what each is (in the first paragraph).. They are different measures of an 'average' - there is no arithmetic relationship between them - that depends on what you are measuring - sometimes they can all be the same, sometimes widely different.. 'Arithmetic mean' is the one most closely similar to someones concept of an average. The 'mode' can be used on non-numerical sets of data - such as popular names.
Do a search outside wikipedia should give less technical descriptions of these things and their uses..87.102.114.215 (talk) 14:56, 14 February 2008 (UTC)
You can find out something about the skewness of a distribution by comparing the mean, median and mode - I can't remember the details, but that article gives some explanation of it (most of it rather technical, though). --Tango (talk) 17:06, 14 February 2008 (UTC)
Mean, median and mode are central tendency measures: They show where data concentration occur.
The mean is the most frequently used of the three concepts. The fact of its taking into account every point of information is its great advantage, but it also could be a drawback, especially whenever outliers are present. In such cases, usually the median is used as a central tendency measure.
As Tango points out, mean, median and mode relate to skewness. Right-skewed distributions have a mean greater in value than the median, and viceversa for negative-skewed distributions. Usually the mode is located in such a way that the median stands between mode and mean. The article on skewness has a section that deals with Pearson skewness coefficients, measures which elaborate on this fact. Pallida  Mors 18:11, 14 February 2008 (UTC)
The article Mode (statistics) also gives a comparison of mean, median and mode.  --Lambiam 18:41, 14 February 2008 (UTC)

[edit] On Average

How many different measures of the "average" of a set of data are there? I know about the mean, median, and mode, are there any others? --Carnildo (talk) 21:23, 14 February 2008 (UTC)

Average has a list of measures of central tendency. --PalaceGuard008 (Talk) 21:28, 14 February 2008 (UTC)