Area of Two Functions: Find the Answer!

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SUMMARY

The discussion focuses on calculating the area between the curves defined by the equations \(y^2 = x + 1\) and \(y = -x + 1\). The user initially attempted to find the area using the integral \(\int_0^3 \sqrt{x+1} - (-x+1) \, dx\), which resulted in an incorrect value of \(\frac{37}{6}\). After guidance, the user correctly identified the need to integrate the area enclosed by the curves from \(x = -2\) to \(x = 1\) using the integral \(\int_{-2}^1 (1 - y) - (y^2 - 1) \, dy\), yielding the correct area of \(\frac{9}{2}\).

PREREQUISITES
  • Understanding of integral calculus, specifically area under curves.
  • Familiarity with the equations of parabolas and linear functions.
  • Knowledge of how to set up and evaluate definite integrals.
  • Ability to visualize functions and their intersections graphically.
NEXT STEPS
  • Study the method of finding areas between curves using definite integrals.
  • Learn about the graphical representation of functions and their intersections.
  • Explore the concept of changing variables in integrals for more complex areas.
  • Review examples of integrating functions that are bounded by multiple curves.
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Students and educators in calculus, mathematicians interested in area calculations, and anyone looking to improve their understanding of integration techniques involving multiple functions.

Petrus
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Hello MHB,
Decide area of the two function
$$y^2=x+1$$ and $$y=-x+1$$

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

Regards,
$$|\pi\rangle$$
 
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Re: area of two function

Petrus said:
Hello MHB,
Decide area of the two function
$$y^2=x+1$$ and $$y=-x+1$$

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

Regards,
$$|\pi\rangle$$

Hi Petrus!

You're integrating the wrong area.
Can you make a drawing?
That should clear things up.
 
Re: area of two function

I like Serena said:
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 $$x=y^2-1$$ and $$x=1-y$$ so we got $$x=-2$$ and $$x=1$$ we got
$$\int_{-2}^11-y-(y^2-1) = \frac{9}{2}$$ which is correct. I guess my x limit is wrong and I don't see how

Regards,
$$|\pi\rangle$$
 
Re: area of two function

Petrus said:
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 $$x=y^2-1$$ and $$x=1-y$$ so we got $$x=-2$$ and $$x=1$$ we got
$$\int_{-2}^11-y-(y^2-1) = \frac{9}{2}$$ which is correct. I guess my x limit is wrong and I don't see how

Regards,
$$|\pi\rangle$$

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.
 

Attachments

  • area_between.png
    area_between.png
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Re: area of two function

I like Serena said:
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.
 
Re: area of two function

Petrus said:
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?
 
Re: area of two function

I like Serena said:
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
35jw7cg.png

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,
$$|\pi\rangle$$
 
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).
 

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