Evaluate this integral from its standard from ?

[osker246]

Homework Statement

Evaluate the integral; [itex]\frac{2}{L}[/itex][itex]\int[/itex]sin²([itex]\frac{\pi*x}{L}[/itex])dx from [itex]\frac{2a}{3}[/itex] to [itex]\frac{a}{3}[/itex]. Where L is a constant, using the standard form [itex]\int[/itex]sin²(ax)dx=[itex]\frac{x}{2}[/itex]-[itex]\frac{1}{4a}[/itex]sin(2ax)+C, where a and C are constants.

The Attempt at a Solution

Ok, so I am taking a physical chemistry and my first homework assignment is more of a review on my math skills. I know how to intergrate but I have never done a problem asking to evaluate an integral using a standard form. So I am not really sure what to do.

Do I evaluate this?

([itex]\frac{x}{2}[/itex]-[itex]\frac{1}{4a}[/itex]sin(2ax)+C)[itex]^{2a/3}_{a/3}[/itex]

Any help is appreciated, thank you!

 

[Pengwuino]

Kinda. First you need to identify the various constants in your problem. For example, your sine function has as the argument [itex]{{\pi x}\over{L}}[/itex] whereas the "standard form is[itex]{{ax}}[/itex] so you need to identify [itex]a = {{\pi}\over{L}}[/itex] so that in your answer, instead of 'a', you have the relevant constant with your problem.

Also, whatever your answer is, remember to multiply by 2/L.

Finally, the +C is for an indefinite integral. You have the limits of integration so what you're doing is evaluating [itex]{{x}\over{2}}-{{1}\over{4a}}sin(2ax)[/itex] at your limits 2a/3 and a/3.

One thing that might catch you is that the 'a' in your problem must have something to do with whatever your problem is, but the 'a' in the "standard form" is NOT that 'a'. It's simply meant to show a constant multiplying the 'x'

 

[TylerH]

Those parentheses should be square brackets.

In general, [itex]\left[ f(x) \right]^b_a = f(b) - f(a)[/itex].

Specifically, [itex]\left[\frac{x}{2}-\frac{sin(2ax)}{4a}+C\right]^{\frac{2a}{3}}_{\frac{a}{3}} = \frac{\frac{2a}{3}}{2}-\frac{sin(2a\frac{2a}{3})}{4a}+C - \frac{\frac{a}{3}}{2}-\frac{sin(2a\frac{a}{3})}{4a}+C[/itex]. Notice that C always goes away, so there's no need to write it.

 

[osker246]

Kinda. First you need to identify the various constants in your problem. For example, your sine function has as the argument [itex]{{\pi x}\over{L}}[/itex] whereas the "standard form is[itex]{{ax}}[/itex] so you need to identify [itex]a = {{\pi}\over{L}}[/itex] so that in your answer, instead of 'a', you have the relevant constant with your problem.

Also, whatever your answer is, remember to multiply by 2/L.

Finally, the +C is for an indefinite integral. You have the limits of integration so what you're doing is evaluating [itex]{{x}\over{2}}-{{1}\over{4a}}sin(2ax)[/itex] at your limits 2a/3 and a/3.

One thing that might catch you is that the 'a' in your problem must have something to do with whatever your problem is, but the 'a' in the "standard form" is NOT that 'a'. It's simply meant to show a constant multiplying the 'x'

So I think I got it. So recognize that a=pi/L and plug it and evaluate the integral with the limits of intergration I have. Its been way to long since calc II, I vaguely remember the standard form stuff now. If memory serves me right, the standard form eqns. would of been in the back of the text. Thanks for clearing that up for me.

Those parentheses should be square brackets.

In general, [itex]\left[ f(x) \right]^b_a = f(b) - f(a)[/itex].

Specifically, [itex]\left[\frac{x}{2}-\frac{sin(2ax)}{4a}+C\right]^{\frac{2a}{3}}_{\frac{a}{3}} = \frac{\frac{2a}{3}}{2}-\frac{sin(2a\frac{2a}{3})}{4a}+C - \frac{\frac{a}{3}}{2}-\frac{sin(2a\frac{a}{3})}{4a}+C[/itex]. Notice that C always goes away, so there's no need to write it.

Sorry, I didn't think about using brackets. I spent like 10 min trying to find all the proper ways of entering my equations, I just got tired and took the easy way out. Thanks for your help too, I appreciate it tons.