# Finding the area

#### shamieh

##### Active member
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 $$\displaystyle \int ^1_0 Right - left \, dx$$

#### Ackbach

##### Indicium Physicus
Staff member
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?

#### shamieh

##### Active member
Top - Bottom ?

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

#### MarkFL

Staff member
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.

#### Prove It

##### Well-known member
MHB Math Helper
WORD OF GOD:

Always draw a diagram.