Area under inverse graphs (1 Viewer)

[alez]

In my ext test, we had to prove the area under sin^-1(x/3) + pi/2 =3pi
We were supposed to draw a rectangle or whatever and do half the area of that but i just changed it to sin(x/3) + pi/2 and integrated that, and it worked
But did it just work for this instance or will it work in others?

 

[tommykins]

回复: Area under inverse graphs

In my ext test, we had to prove the area under sin^-1(x/3) + pi/2 =3pi
We were supposed to draw a rectangle or whatever and do half the area of that but i just changed it to sin(x/3) + pi/2 and integrated that, and it worked
But did it just work for this instance or will it work in others?

You cannot just randomly change a graph, you will not score any marks for it.

The graph of asin(x/3) + pi/2 is different from sin(x/3) + pi/2, you've just changed the question completely.

 

[conics2008]

Re: 回复: Area under inverse graphs

what you did was correct buddy..

because if you look at the arcsin(x) and the normal sin(x) their basically the same but your y values become your x values..

what you did was the proper way to solve it...

tommy thats how you solve it.. he didn't even change the question..

see if

y=arc sin(x/3) finding the area of that you would need to find the yint because these were your old x intercept..

therefore siny=x/3 >> x=3siny <<<< integrate this...

 

[tommykins]

回复: Re: 回复: Area under inverse graphs

what you did was correct buddy..

because if you look at the arcsin(x) and the normal sin(x) their basically the same but your y values become your x values..

what you did was the proper way to solve it...

tommy thats how you solve it.. he didn't even change the question..

see if

y=arc sin(x/3) finding the area of that you would need to find the yint because these were your old x intercept..

therefore siny=x/3 >> x=3siny <<<< integrate this...

I know that they're similar, but he said he changed it sin(x/3) and not sin(y/3) which got me confused.

Your response was better than mine, kudos.