Area of figure defined by function

[Loki123]

In the problem where it says function it's supposed to be figure.

I have never encountered sin in this context. Where did I go wrong? My answer is way off.

 

[lex]

You have integrated [imath]\sin{\tfrac{\pi x}{2}}[/imath] incorrectly.

Try differentiating [imath]\tfrac{\pi}{2}\cos{\tfrac{\pi x}{2}}[/imath] to see.

 

[Loki123]

You have integrated [imath]\sin{\tfrac{\pi x}{2}}[/imath] incorrectly.

Try differentiating [imath]\tfrac{\pi}{2}\cos{\tfrac{\pi x}{2}}[/imath] to see.

I think I fixed that.

 

[BigBeachBanana]

I think I fixed that. View attachment 32243

Still wrong. Since you can't see it right away, use u-subtitution.
Let [imath]u=\frac{\pi x}{2} \implies du=? dx[/imath]. Continue.

 

[Loki123]

Still wrong. Since you can't see it right away, use u-subtitution.
Let [imath]u=\frac{\pi x}{2} \implies du=? dx[/imath]. Continue.

Oh I get it. x^2/2. I accidentally used derivative way.

 

[Steven G]

You commuted the derivative of sin(pix/2) when you were asked to integrate it.

If u=pix/2, then dx =???

Not sure what you are saying about x^2/2(???)

 

[Loki123]

You commuted the derivative of sin(pix/2) when you were asked to integrate it.

If u=pix/2, then dx =???

Not sure what you are saying about x^2/2(???)

it's supposed to be Pix^2/4

 

[Loki123]

Still wrong. Since you can't see it right away, use u-subtitution.
Let [imath]u=\frac{\pi x}{2} \implies du=? dx[/imath]. Continue.

du= Pi/2 dx

 

[Steven G]

du= Pi/2 dx

So dx = (2/pi)du. Very good.

 

[Loki123]

So dx = (2/pi)du. Very good.

I get it, I was supposed to change dx to du, instead of just changing x to u. I confused integral and derivative sometimes.

 

[Steven G]

it's supposed to be Pix^2/4

What is supposed to be pix^2/4??

 

[Loki123]

What is supposed to be pix^2/4??

unnecessary derivative of pix/2

 

[Loki123]

unnecessary derivative of pix/2

actually not even that, i just got confused.