Integral and Area Bounded by Curves

[TwistedNerve]

Hello. I have two math problems.

Here is the first.

Find the area of the region bounded by \(\displaystyle x^{2}+y^{2}=1\), \(\displaystyle y=0\) and \(\displaystyle y=x^{2}\). I needed to only be concerned with Quadrant I

I began by sketching a picture and saw I needed to add two integrals. The first would be \(\displaystyle \int_{0}^{a}x^{2} dx\) where a= intersection of \(\displaystyle x^{2}+y^{2}=1\) and \(\displaystyle y=x^{2}\). The second would be \(\displaystyle \int_{a}^{1}{\sqrt{-x+1}}\ dx\). Soo I have two questions. First, did I set this up right? And second, how do I find the point of intersection without using a calculator? I set the two equations equal to each other and tried solving for X but that got me no where.


My second problem is \(\displaystyle \int\frac{x}{\sqrt{x}+1} dx\)

Soo i spent about a half hour on this problem last night (it is only suppose to take 5-10 minutes max). I didnt see any good u substitutions. If you point one out to me Ill be embarassed. I thought of using the table method but I wasnt sure if that would get me anywhere. So I basically just started guessing and checking. The closest I got was \(\displaystyle 2x^2ln({\sqrt{x}+x})\) This answer ends up giving me the right answer + 1 + \(\displaystyle 2ln({\sqrt{x}+x})\). So I thought if I could figure out the integral of that I could just subtract it and get the right answer but I got stuck. I know Im kind of using a brute force method so is there an easier way? Thanks.

 

[tkhunny]

x^2 + y^2 = 1 and y = x^2

Substitute!

y + y^2 = 1 ==> \(\displaystyle y\;=\;\frac{\sqrt{5}-1}{2}\) There is also a negative solution, but we're in the first quadrant, right?

Use this value in either equation to find the corresponding x-value. This is your 'a'.

On the other hand, why do you need two integrals. Use the y-axis!

\(\displaystyle \int_{0}^{\frac{\sqrt{5}-1}{2}}\left(\sqrt{1-y^{2}}-\sqrt{y}\right)\;dy\)

For the second, why do you think you need to find another expression for this integral? Are you SURE it is the right one? Perhaps a problem was set up badly? Not all integrals that can be expressed easily as integrals also have a simple non-integral expression. This is one such integral expression. Why do you think it can be solved by simple substitution and takes only 10 minutes? I suppose the 10 minutes is okay if you count the time to determine that it can't be done with elementary expressions.

 

[soroban]

Hello, TwistedNerve!

You're doing fine with the first one . . .
Well, you left off a square in the second integral.

Find the area of the region bounded by \(\displaystyle x^2+y^2\:=\:1,\;\;y=0,\;\;y=x^2\)

I began by sketching a picture. Good! . and saw I needed to add two integrals.
\(\displaystyle \text{The first would be: }\;\int_0^ax^2\,dx\;\text{ where }a\text{ = intersection of the graphs.}\)

\(\displaystyle \text{The second would be: }\;\int_a^1\sqrt{1-x^2}\,dx\).

So I have two questions.
First, did I set this up right? . . . . yes!
And second, how do I find the point of intersection without using a calculator?
I set the two equations equal to each other and tried solving for X but that got me nowhere.

We have: .\(\displaystyle y = x^2\) and \(\displaystyle x^2 + y^2\:=\:1\)

Substitute the first into the second: .\(\displaystyle y + y^2 \:=\:1\quad\Rightarrow\quad y^2 + y - 1 \:=\:0\)

\(\displaystyle \text{Quadratic Formula: }\;y \:=\:\frac{\sqrt{5} - 1}{2}\)

\(\displaystyle \text{Substitute into the second equation and solve for }x:\;\;x \:=\:\sqrt{\sqrt{5}-2}\)

\(\displaystyle \text{My second problem is: }\;\int\frac{x}{\sqrt{x}+1} dx\)

Let: \(\displaystyle u = \sqrt{x}\quad\Rightarrow\quad x = u^2\quad\Rightarrow\quad dx = 2u\,du\)

\(\displaystyle \text{Substitute: } \;\int\frac{u^2}{u+1}(2u\,du) \;=\;2\int\frac{u^3}{u+1}\,du\)

\(\displaystyle \text{Long division: }\;2\int\left(u^2 - u + 1 + \frac{1}{u+1}\right)\,du\)