Evaluate this integral from its standard from ?

In summary, the integral can be evaluated by identifying the relevant constants, recognizing that a = pi/L, and plugging it into the standard form equation. The final result will not include the constant C. The parentheses in the equations should be square brackets for proper notation.
  • #1
osker246
35
0

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!
 
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  • #2
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'
 
  • #3
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.
 
Last edited:
  • #4
Pengwuino said:
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.

TylerH said:
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.
 

Related to Evaluate this integral from its standard from ?

1. What is an integral?

An integral is a mathematical concept used in calculus to find the area under a curve or the accumulation of a quantity over an interval. It is represented by the symbol ∫ and has two components: the integrand (the function being integrated) and the limits of integration.

2. What is the standard form of an integral?

The standard form of an integral is ∫f(x)dx, where f(x) is the integrand, and dx represents the variable of integration. This form is used to represent the indefinite integral or antiderivative of a function. For definite integrals, the standard form is ∫f(x)dx from a to b, where a and b are the lower and upper limits of integration, respectively.

3. How do you evaluate an integral?

To evaluate an integral, you can use various techniques such as substitution, integration by parts, or trigonometric substitution. The goal is to manipulate the integrand into a form that can be easily integrated, and then use the fundamental theorem of calculus to find the solution. It is also essential to pay attention to any given limits of integration and apply them correctly.

4. What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that the definite integral of a function can be evaluated by finding its antiderivative and evaluating it at the upper and lower limits of integration. In other words, it connects the concepts of differentiation and integration and allows us to easily evaluate definite integrals without having to use the limit definition.

5. What are some real-life applications of integrals?

Integrals have various practical applications in fields such as physics, engineering, economics, and statistics. For example, they can be used to calculate the work done by a force, find the center of mass of an object, determine the area under a velocity-time graph, and calculate the total revenue from a demand function. They are also used in machine learning and data analysis to find the area under a curve representing a probability distribution.

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