Find the number a such that the line x = a divides the region bounded by the curves x = y2 − 1 and the y-axis into 2 regions with equal area. Give your answer correct to 3 decimal places.

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Find the number a such that the line x = a divides the region bounded by the curves x = y2 − 1 and the y-axis into 2 regions with equal area. Give your answer correct to 3 decimal places.

Mathematics
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At first the full are that we are talking about: The boundaries are: \(\large\color{black}{ \displaystyle x=y^2-1 }\) \(\large\color{black}{ \displaystyle x=0 }\) ------------------------------------ |dw:1437950043784:dw|
the full area*
So far we can tell that -1

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rearrangement \(x=y^2-1\) \(y=\sqrt{x+1}\) \(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}\left(\sqrt{x+1}\right)~dx}\) is the area of the full region. Now, we are splitting it into two regions that stop and start (respectively) at x=a. \(\large\color{slate}{\displaystyle\int\limits_{-1}^{a}\left(\sqrt{x+1}\right)~dx=\int\limits_{a}^{0}\left(\sqrt{x+1}\right)~dx}\)
(and initially, want to say that I mean that a is between x=-1 and x=0) any questions/
?
I got this far but cant seem to get a good answer I got an answer of -.023
\(\large \displaystyle \int\limits_{-1}^{a}\left(\sqrt{x+1}\right)~dx=\int\limits_{a}^{0}\left(\sqrt{x+1}\right)~dx\) \(\large \displaystyle \left.(x+1)^{3/2}~~\right|^{a}_{-1}=\left.(x+1)^{3/2}~\right|^{0}_{a}\) \(\large \displaystyle (a+1)^{3/2}-0^{3/2}=(1)^{3/2}- (a+1)^{3/2}\) \(\large \displaystyle (a+1)^{3/2}=1- (a+1)^{3/2}\)
\(\large \displaystyle2 (a+1)^{3/2}=1\) \(\large \displaystyle(a+1)^{3/2}=1/2\) \(\large \displaystyle a+1=(1/2)^{2/3}\) \(\large \displaystyle a=1/\sqrt[3]{4}-1\) \(a\approx -0.37\)
this is what i got
we can do a taylor polynomial approximation for \(\sqrt[3]{4}\) if you like:)
thank you very much!
questions about the problem?
no your explanation answers all questions thank you very much
Alright. yw

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