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Area of two function

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
Decide area of the two function
\(\displaystyle y^2=x+1\) and \(\displaystyle y=-x+1\)

progress:
first I equal them to see where they intercept and get \(\displaystyle x=0\) and \(\displaystyle x=3\)
then I do integrate
\(\displaystyle \int_0^3 \sqrt{x+1}-(-x+1) = \frac{37}{6}\) which is not correct answer, What I am doing wrong?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Re: area of two function

Hello MHB,
Decide area of the two function
\(\displaystyle y^2=x+1\) and \(\displaystyle y=-x+1\)

progress:
first I equal them to see where they intercept and get \(\displaystyle x=0\) and \(\displaystyle x=3\)
then I do integrate
\(\displaystyle \int_0^3 \sqrt{x+1}-(-x+1) = \frac{37}{6}\) which is not correct answer, What I am doing wrong?

Regards,
\(\displaystyle |\pi\rangle\)
Hi Petrus!

You're integrating the wrong area.
Can you make a drawing?
That should clear things up.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: area of two function

Hi Petrus!

You're integrating the wrong area.
Can you make a drawing?
That should clear things up.
Hello I like Serena,
does not make clear when I draw it but I don't see what I am doing wrong:S
When I make it \(\displaystyle x=y^2-1\) and \(\displaystyle x=1-y\) so we got \(\displaystyle x=-2\) and \(\displaystyle x=1\) we got
\(\displaystyle \int_{-2}^11-y-(y^2-1) = \frac{9}{2}\) which is correct. I guess my x limit is wrong and I don't see how

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Re: area of two function

Hello I like Serena,
does not make clear when I draw it but I don't see what I am doing wrong:S
When I make it \(\displaystyle x=y^2-1\) and \(\displaystyle x=1-y\) so we got \(\displaystyle x=-2\) and \(\displaystyle x=1\) we got
\(\displaystyle \int_{-2}^11-y-(y^2-1) = \frac{9}{2}\) which is correct. I guess my x limit is wrong and I don't see how

Regards,
\(\displaystyle |\pi\rangle\)
Ok. Let me put in a picture then.

area_between.png

You integrated the upper-right area in the parabola, which is also bounded by x=3.
But the lower-left area inside the parabola is needed.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: area of two function

Ok. Let me put in a picture then.

View attachment 884

You integrated the upper-right area in the parabola, which is also bounded by x=3.
But the lower-left area inside the parabola is needed.
Hello I like Serena,
I still find it hard to see if I calculate the correct area now when I think about it.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Re: area of two function

Hello I like Serena,
I still find it hard to see if I calculate the correct area now when I think about it.
Do you understand that you need the bottom-left area, which is completely enclosed by the 2 curves?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: area of two function

Do you understand that you need the bottom-left area, which is completely enclosed by the 2 curves?
Hello I like Serena,
I understand that we want this area

Edit: I think I understand now, I think I have confused myself.. I will be back if I still got problem, Thanks once again for the help I like Serena!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Re: area of two function

Good.

In your initial integral, you took $y=\sqrt{x+1}$, which is the top half of the parabola.
And you subtracted $y=-x+1$, which is the line, that now counts as the bottom.
Then you did so for each x between 0 and 3.
This is the upper-right part inside the parabola.

In your second, correct, attempt, you took $x=1-y$, which is the line, which now counts as the boundary on the right.
And you subtracted $x=y^2-1$, which is the left part of the parabola.
This is the proper bottom-left part inside the parabola.

See for instance this page with more pictures (and explanations).