## anonymous one year ago 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.

1. SolomonZelman

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|

2. SolomonZelman

the full area*

3. SolomonZelman

So far we can tell that -1<a<0.

4. SolomonZelman

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. $$\large\color{slate}{\displaystyle\int\limits_{-1}^{a}\left(\sqrt{x+1}\right)~dx=\int\limits_{a}^{0}\left(\sqrt{x+1}\right)~dx}$$

5. SolomonZelman

(and initially, want to say that I mean that a is between x=-1 and x=0) any questions/

6. SolomonZelman

?

7. anonymous

I got this far but cant seem to get a good answer I got an answer of -.023

8. SolomonZelman

$$\large \displaystyle \int\limits_{-1}^{a}\left(\sqrt{x+1}\right)~dx=\int\limits_{a}^{0}\left(\sqrt{x+1}\right)~dx$$ $$\large \displaystyle \left.(x+1)^{3/2}~~\right|^{a}_{-1}=\left.(x+1)^{3/2}~\right|^{0}_{a}$$ $$\large \displaystyle (a+1)^{3/2}-0^{3/2}=(1)^{3/2}- (a+1)^{3/2}$$ $$\large \displaystyle (a+1)^{3/2}=1- (a+1)^{3/2}$$

9. SolomonZelman

$$\large \displaystyle2 (a+1)^{3/2}=1$$ $$\large \displaystyle(a+1)^{3/2}=1/2$$ $$\large \displaystyle a+1=(1/2)^{2/3}$$ $$\large \displaystyle a=1/\sqrt[3]{4}-1$$ $$a\approx -0.37$$

10. SolomonZelman

this is what i got

11. SolomonZelman

we can do a taylor polynomial approximation for $$\sqrt[3]{4}$$ if you like:)

12. anonymous

thank you very much!

13. SolomonZelman

14. anonymous