- Thread starter
- #1

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

- Thread starter shamieh
- Start date

- Thread starter
- #1

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

- Admin
- #2

- Jan 26, 2012

- 4,198

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?

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

- Thread starter
- #3

- Admin
- #4

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.Top - Bottom ?

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