Wikipedia:Reference desk/Archives/Mathematics/2007 December 29

= December 29 =

Integration (2)
I have two equations, $$ y_1=5+x-x^2$$ and $$y_2=x+4$$, and I have to find the area between them, bounded by their points of intersection. Now I can do this by integrating $$ y_1$$ with between the points of intersection, then integrating $$ y_2$$ between the points of intersection and subtracting the second from the first. Could it be done by integrating $$ x^2-1$$, the equation you get when you equate the first two and set one side equal to zero, between the points of intersection? 172.203.54.211 (talk) 17:21, 29 December 2007 (UTC)


 * So, you're asking if $$\int_a^b y_1 dx - \int_a^b y_2 dx = \int_a^b(y_1-y_2)dx$$ ? The answer is yes, because integration is a linear operator. Just be sure to keep your signs straight. —Keenan Pepper 17:33, 29 December 2007 (UTC)


 * What you say about signs is interesting because, depending on how you re-arrangem you can get either $$ x^2-1$$ or $$1-x^2$$. How do you know which is the one you want? 172.203.54.211 (talk) 17:38, 29 December 2007 (UTC)


 * Just sketch the graph of the equations. The one above the other will be positive. --Taraborn (talk) 21:09, 29 December 2007 (UTC)


 * (After two edit conflicts!) Both procedures are right. In plain words, "the integral of a difference is the difference of the integrals".


 * Remember that definite integrals calculate algebraic areas, rather than geometric surface. So, in your case, since you are substracting the function with greater values in the interval, your last choice will produce a negative value; its corresponding absolute value gives the wanted area. Pallida  Mors  17:40, 29 December 2007 (UTC)

Splitting a number into two by percentage
Is it possible to split a number into two numbers in such a way that the first number is a certain percentage of the second number. For example, splitting 110 using 10% would result in 10 and 100. Is there a way to do this for any number? --24.26.43.4 (talk) 17:41, 29 December 2007 (UTC)
 * Well, say you call the initial number n, these split numbers a and b, and a factor p. You want $$a+b=n$$ where $$a=bp$$.  Finding equations for a and b is then quite easy.  x42bn6 Talk Mess  18:02, 29 December 2007 (UTC)
 * Hang on a minute. 10% of 110 isn't 10, it's 11!  If you want to find x% of 110, you multiply 110 by $$ (x / 100) $$.  So 10% of 110 is $$ 110 \times (10/100) = 11$$. Silverfish70 (talk) 23:27, 29 December 2007 (UTC)
 * The OP said, "the first number is a certain percentage of the second number", not of the original number. 10% of 100 is indeed 10. -- Meni Rosenfeld (talk) 23:50, 29 December 2007 (UTC)
 * I misread - sorry. Then we want to write 110 as $$a+b$$ where b is x% of a.  So $$b = (x/100) \times a$$, and we've got $$ a + (x/100)a = 110$$.  Since we know x, we can now find a  (and b). Doing this in the example given leads to $$1.1 a =110$$ so a is 100 and b is 10, as the OP said.  Silverfish70 (talk) 16:30, 30 December 2007 (UTC)