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

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SUMMARY

The area bounded by the curves \(y = x^2\) and \(y = 2x - x^2\) is determined by solving the equation \(x^2 = 2x - x^2\), leading to the factorization \(2x(x - 1)\), which gives the intersection point \(x = 1\). The area can be calculated using the integral \(\int_0^1 (2x - x^2 - x^2) \, dx\), simplifying to \(\int_0^1 (2x - 2x^2) \, dx\). It is crucial to identify the top and bottom functions correctly for integration, which can be verified by evaluating the functions at a point within the integration limits. Drawing a diagram is recommended for clarity.

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shamieh
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Find the area bounded by the curves $$y = x^2[/math] and [math]y = 2x - x^2$$so
$$
x^2 = 2x - x^2$$
$$
2x - x^2 - x^2 = 2x - 2x^2$$

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

So the $$\int ^1_0 Right - left \, dx$$
 
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shamieh said:
Find the area bounded by the curves $$y = x^2[/math] and [math]y = 2x - x^2$$so
$$
x^2 = 2x - x^2$$
$$
2x - x^2 - x^2 = 2x - 2x^2$$

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

So the $$\int ^1_0 Right - left \, dx$$

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?
 
Top - Bottom ?So would i get $$\int^1_0 x^2 - (2x -x ^2) dx$$
 
shamieh said:
Top - Bottom ?So would i get $$\int^1_0 x^2 - (2x -x ^2) dx$$

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.
 
WORD OF GOD:

Always draw a diagram.
 

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