• Calculus help?
    9 replies, posted
We just learned u-substitution for integration in my calculus 1 class. This problem has me totally stumped.. integrate x*sqrt(1-x^4) dx from x=0 to x=1, using the substitution u=x^2, and interpreting the resulting definite integral as an area. I know that du/dx would be 2x dx (duh.) I just can't figure out how to integrate (1/2) (integral from 0 to 1) [ (1-u^2)^(1/2) ] du Can any of you math wizards help me?
hang on i'll work it out on paper
We really need a "Help at School" forum.
ok, i can see where you've gone wrong, let me upload pictures of the correct working [editline]29th April 2013[/editline] but now ive gone wrong, so fuck it i give up, sorry
I figured it out. If we use u = x^2, then it turns into the area of a circle equation. Durr.
That was one of today's examples, but I forgot it as well. I was getting confused when Wolfram Alpha was suggesting it had something to do with inverse sin.
(integral from 0 to 1) [ x*(1-x^4)^(1/2) ] dx u = x^2; du/dx = 2x dx (1/2) (integral from 0 to 1) [ 2x*(1-x^4)^(1/2) ] dx (1/2) (integral from 0 to 1) [ (1-u^2)^(1/2) ] du (1/2)(1/4)(pi*1^2) == (pi/8) This is because we're taking the area of 1/4th of a circle of radius 1, (A=pi*r^2) , then multiplying that by the 1/2 that canceled out with the 2 in du/dx.
[QUOTE=Hypershadsy;40469163]That was one of today's examples, but I forgot it as well. I was getting confused when Wolfram Alpha was suggesting it had something to do with inverse sin.[/QUOTE] It does. This integral can be solved with a simple u substitution, but when it comes to a piece of a circle being bound by two vertical lines (which doesn't happen to divide it into quarters or halves), the arcsine will give an antiderivative that works for way more integrals.
You can put this into Wolfram Alpha and it will solve it for you plus give you step by step instructions on how to solve it through various methods.
I don't trust Stephen Wolfram with my email.
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