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

[shamieh]

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

 

[Ackbach]

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?

 

[shamieh]

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

 

[MarkFL]

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.

 

[Prove It]

WORD OF GOD:

Always draw a diagram.