Calculus 1 or AP Calculus
Guys please help me, I give medals

- anonymous

Calculus 1 or AP Calculus
Guys please help me, I give medals

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- anonymous

##### 1 Attachment

- anonymous

@jim_thompson5910

- anonymous

@DanJS

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- anonymous

I know the total area bounded by the curves is 10.6666.

- anonymous

|dw:1449188701513:dw|

- jim_thompson5910

Yes,
\[\Large \int_{-2}^{2}(4-x^2)dx = \frac{32}{3} \approx 10.66667\]

- anonymous

|dw:1449188844607:dw|

- anonymous

but I don't know how to find the line that would cut the area into two

- anonymous

i know x=0 would, but it has to be y=b

- 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

okay

- anonymous

but how did you get the limits -b^(1/2) and b^1/2)?

- jim_thompson5910

y = x^2
b = x^2 .... plug in y = b
x^2 = b
x = sqrt(b) or x = -sqrt(b)

- anonymous

oh I see, so 5.333 has to be the value of b right?

- jim_thompson5910

that's half of the area you got before
that is not the value of b

- anonymous

oh okay, .... so now how can I find the value of b?

- jim_thompson5910

Let
\[\Large g(x) = \int(b-x^2)dx\]

- 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

now compute \[\Large g(\sqrt{b})-g(-\sqrt{b})\] and tell me what you get

- anonymous

I still don't understand what to do from here.

- jim_thompson5910

what did you get when you computed \[\Large g(\sqrt{b})-g(-\sqrt{b})\]

- anonymous

I got b(b^(1/2) - ((b^(1/2)) ^3)/3 -[ b(-b^(1/2)) - ((-b^(1/2))^3)/3) ]

- anonymous

I got that and I know I did it wrong

- 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

Im so confused right now, can you tell me how did you get 2/3?

- jim_thompson5910

did you see how I got \[\Large g(x) = bx - \frac{x^3}{3}+C\]

- anonymous

oh okay I see, you took the antiderivative

- jim_thompson5910

yes then plugged in x = sqrt(b) and x = -sqrt(b)

- jim_thompson5910

then subtracted

- anonymous

so that means that 10.666= [4/3 b^ (3/2)]?

- jim_thompson5910

not quite

- 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

\[\Large g(\sqrt{b})-g(-\sqrt{b})=\frac{16}{3}\]
\[\Large \frac{4}{3}b^{3/2}=\frac{16}{3}\]
now solve for b

- anonymous

8

- anonymous

wait

- jim_thompson5910

b isn't a whole number

- anonymous

2.51984

- 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

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

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- 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

hint: the area you want is shown in blue (see attached)

##### 1 Attachment

- 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

so basically like this

##### 1 Attachment

- 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

not x^2 - 1, just x^2

- anonymous

okay, so B would be the answer correct?

- jim_thompson5910

yes I agree

##### 1 Attachment

- anonymous

Oh my god thank you so much.

- jim_thompson5910

glad to be of help

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