Finding Area Bounded by x^2 & 2x - x^2

  • MHB
  • Thread starter shamieh
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In summary, to find the area bounded by the curves y = x^2 and y = 2x - x^2, we can set the equations equal to each other and solve for x, which gives us the bounds of integration. When integrating with respect to x, the top function is subtracted by the bottom function. It is recommended to draw a diagram to determine the correct order of the functions when integrating.
  • #1
shamieh
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Find the area bounded by the curves \(\displaystyle y = x^2\) and \(\displaystyle y = 2x - x^2\)so
\(\displaystyle
x^2 = 2x - x^2\)
\(\displaystyle
2x - x^2 - x^2 = 2x - 2x^2\)

So then would I factor out a 2 and get
\(\displaystyle
2x(x - 1)\)
\(\displaystyle
x = 1\)

So the \(\displaystyle \int ^1_0 Right - left \, dx\)
 
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  • #2
shamieh said:
Find the area bounded by the curves \(\displaystyle y = x^2\) and \(\displaystyle y = 2x - x^2\)so
\(\displaystyle
x^2 = 2x - x^2\)
\(\displaystyle
2x - x^2 - x^2 = 2x - 2x^2\)

So then would I factor out a 2 and get
\(\displaystyle
2x(x - 1)\)
\(\displaystyle
x = 1\)

So the \(\displaystyle \int ^1_0 Right - left \, dx\)

The "Right" and "Left" is if you're doing an integral with respect to $y$. But you're integrating w.r.t. $x$. So it's what minus what?
 
  • #3
Top - Bottom ?So would i get \(\displaystyle \int^1_0 x^2 - (2x -x ^2) dx\)
 
  • #4
shamieh said:
Top - Bottom ?So would i get \(\displaystyle \int^1_0 x^2 - (2x -x ^2) dx\)

Yes, it is top minus bottom, but are you sure you have chosen correctly with regards to which is top and which is bottom? If you are unsure, pick an $x$-value inside the limits of integration and evaluate both functions to see which gives the greater value.
 
  • #5
WORD OF GOD:

Always draw a diagram.
 

FAQ: Finding Area Bounded by x^2 & 2x - x^2

What is the formula for finding the area bounded by x^2 and 2x - x^2?

The formula for finding the area bounded by two functions is given by the definite integral of the difference between the two functions. In this case, it would be ∫(2x - x^2 - x^2) dx.

How do I know which function to use as the upper and lower boundaries when finding the area?

The upper and lower boundaries are determined by the points where the two functions intersect. In this case, the two functions intersect at x=0 and x=2. The function 2x - x^2 is above x^2 between these two points, so it would be the upper boundary, while x^2 would be the lower boundary.

Can I use any method other than integration to find the area bounded by x^2 and 2x - x^2?

No, integration is the only method for finding the area bounded by two functions. However, there are different techniques for evaluating integrals such as u-substitution or integration by parts.

Is it possible for the area bounded by x^2 and 2x - x^2 to be negative?

Yes, it is possible for the area to be negative if the two functions intersect in a way that the function 2x - x^2 is below x^2 between the points of intersection. In this case, the area would be calculated as a negative value.

How can I visualize the area bounded by x^2 and 2x - x^2?

You can visualize the area by graphing the two functions on the same coordinate plane and shading in the area between them. This will give you a visual representation of the bounded area.

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