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Finding the area

shamieh

Active member
Sep 13, 2013
539
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
Jan 26, 2012
4,197
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
Sep 13, 2013
539
Top - Bottom ?


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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 26, 2012
1,404
WORD OF GOD:

Always draw a diagram.