anonymous
  • anonymous
Calculus 1 or AP Calculus Guys please help me, I give medals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
@DanJS

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anonymous
  • anonymous
I know the total area bounded by the curves is 10.6666.
anonymous
  • anonymous
|dw:1449188701513:dw|
jim_thompson5910
  • jim_thompson5910
Yes, \[\Large \int_{-2}^{2}(4-x^2)dx = \frac{32}{3} \approx 10.66667\]
anonymous
  • anonymous
|dw:1449188844607:dw|
anonymous
  • anonymous
but I don't know how to find the line that would cut the area into two
anonymous
  • anonymous
i know x=0 would, but it has to be y=b
jim_thompson5910
  • jim_thompson5910
you need to solve for b in this equation \[\Large \int_{-\sqrt{b}}^{\sqrt{b}}(b-x^2)dx = \frac{16}{3}\] I got the limits of integration by solving y = x^2 for x(and I plugged in y = b)
anonymous
  • anonymous
okay
anonymous
  • anonymous
but how did you get the limits -b^(1/2) and b^1/2)?
jim_thompson5910
  • jim_thompson5910
y = x^2 b = x^2 .... plug in y = b x^2 = b x = sqrt(b) or x = -sqrt(b)
anonymous
  • anonymous
oh I see, so 5.333 has to be the value of b right?
jim_thompson5910
  • jim_thompson5910
that's half of the area you got before that is not the value of b
anonymous
  • anonymous
oh okay, .... so now how can I find the value of b?
jim_thompson5910
  • jim_thompson5910
Let \[\Large g(x) = \int(b-x^2)dx\]
jim_thompson5910
  • jim_thompson5910
when you take the indefinite integral of the RHS, you should find that \[\Large g(x) = \int(b-x^2)dx\] \[\Large g(x) = bx - \frac{x^3}{3}+C\]
jim_thompson5910
  • jim_thompson5910
now compute \[\Large g(\sqrt{b})-g(-\sqrt{b})\] and tell me what you get
anonymous
  • anonymous
I still don't understand what to do from here.
jim_thompson5910
  • jim_thompson5910
what did you get when you computed \[\Large g(\sqrt{b})-g(-\sqrt{b})\]
anonymous
  • anonymous
I got b(b^(1/2) - ((b^(1/2)) ^3)/3 -[ b(-b^(1/2)) - ((-b^(1/2))^3)/3) ]
anonymous
  • anonymous
I got that and I know I did it wrong
jim_thompson5910
  • jim_thompson5910
you should find that... \[\Large g(\sqrt{b}) = \frac{2}{3}b^{3/2}+C\] \[\Large g(-\sqrt{b}) = -\frac{2}{3}b^{3/2}+C\] so, \[\Large g(\sqrt{b})-g(-\sqrt{b})=\left(\frac{2}{3}b^{3/2}+C\right)-\left(-\frac{2}{3}b^{3/2}+C\right)\] \[\Large g(\sqrt{b})-g(-\sqrt{b})=\frac{2}{3}b^{3/2}+C+\frac{2}{3}b^{3/2}-C\] \[\Large g(\sqrt{b})-g(-\sqrt{b})=\frac{4}{3}b^{3/2}\]
anonymous
  • anonymous
Im so confused right now, can you tell me how did you get 2/3?
jim_thompson5910
  • jim_thompson5910
did you see how I got \[\Large g(x) = bx - \frac{x^3}{3}+C\]
anonymous
  • anonymous
oh okay I see, you took the antiderivative
jim_thompson5910
  • jim_thompson5910
yes then plugged in x = sqrt(b) and x = -sqrt(b)
jim_thompson5910
  • jim_thompson5910
then subtracted
anonymous
  • anonymous
so that means that 10.666= [4/3 b^ (3/2)]?
jim_thompson5910
  • jim_thompson5910
not quite
jim_thompson5910
  • jim_thompson5910
first off, I like to keep things as fraction whenever possible it turns out that 32/3 = 10.667 approx so the original area is 32/3 take half of this to get 16/3
jim_thompson5910
  • jim_thompson5910
\[\Large g(\sqrt{b})-g(-\sqrt{b})=\frac{16}{3}\] \[\Large \frac{4}{3}b^{3/2}=\frac{16}{3}\] now solve for b
anonymous
  • anonymous
8
anonymous
  • anonymous
wait
jim_thompson5910
  • jim_thompson5910
b isn't a whole number
anonymous
  • anonymous
2.51984
jim_thompson5910
  • jim_thompson5910
I'm getting \[\Large b = 2*\sqrt[3]{2} \approx 2.5198421 \approx 2.520\] so it looks like you got it
anonymous
  • anonymous
yes, I get it now. Thank you so much. Could you help me with another one? like really fast. I get the concept, all I need is to clarify something
anonymous
  • anonymous
anonymous
  • anonymous
Does the bounded area goes from 0 to 2 or from 0 to 1? I did the interceptions and I got -2 and 1, but I got this questions wrong
jim_thompson5910
  • jim_thompson5910
hint: the area you want is shown in blue (see attached)
1 Attachment
jim_thompson5910
  • jim_thompson5910
based on the picture I attached, you'll have to break up the blue area into 2 pieces the first piece will span from A to B the second piece will span from B to C
jim_thompson5910
  • jim_thompson5910
so basically like this
1 Attachment
anonymous
  • anonymous
so for b t c could I do the triangle, and from a to b could I do [x^2 -1] from (0 to 1)?
jim_thompson5910
  • jim_thompson5910
not x^2 - 1, just x^2
anonymous
  • anonymous
okay, so B would be the answer correct?
jim_thompson5910
  • jim_thompson5910
yes I agree
1 Attachment
anonymous
  • anonymous
Oh my god thank you so much.
jim_thompson5910
  • jim_thompson5910
glad to be of help

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