With this inverse trig qu... (1 Viewer)

[VenomP]

Find the area bounded by the curves: y=sin^-1x, y=sin^-1(-x) and the line y=pi/4

I get to the working out stage, and apparently you have to multiply the found area by MINUS two instead of two (i know that you do this because the areas are symmetrical) but why minus two?

Answer: 2-rt2 units squared

 

[xFusion]

you probally dont need to multiply by -2 but since its an area question, the answer is always going to be a positive. Hence that is why you multiplied by -2. I wouldve jst changed my answer to a positive and stated it was an area statement.

 

[undalay]

<a href="http://www.codecogs.com/eqnedit.php?latex== 2\int_{0}^{pi/4} sin x dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?= 2\int_{0}^{pi/4} sin x dx" title="= 2\int_{0}^{pi/4} sin x dx" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==2[-cosx]_{0}^{pi/4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=2[-cosx]_{0}^{pi/4}" title="=2[-cosx]_{0}^{pi/4}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==2(-\sqrt{2}/2 -(-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=2(-\sqrt{2}/2 -(-1)" title="=2(-\sqrt{2}/2 -(-1)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==2(1-\sqrt{2}/2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=2(1-\sqrt{2}/2)" title="=2(1-\sqrt{2}/2)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==2-\sqrt{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=2-\sqrt{2}" title="=2-\sqrt{2}" /></a>