Integration - Change of Variable

[BOAS]

Homework Statement

[/B]
Use integration by substitution to evaluate the integral,

$I = \int^{x}_{x_{0}} (3 + 4t)^{\frac{5}{3}} dt$

Homework Equations

The Attempt at a Solution

I am confused by this question, and think that the limits on the integral might be a typo. Does it make sense for them to be $x, x_{0}$? I think that the question means for them to be $t, t_{0}$ but I'm not sure that it isn't me not understanding something properly.

Ordinarily,

I would make the substitution $u = 3 + 4t$, and say that $du = 4 dt$.

$I = \int^{3+4t}_{3+4t_{0}} (u)^{\frac{5}{3}} \frac{du}{4} = [ \frac{3}{5} \frac{u^{\frac{8}{3}}}{4}]^{3 + 4t}_{3+4t_{0}} = [ \frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}]^{3 + 4t}_{3+4t_{0}}$

$I = (\frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4t_{0})^{\frac{8}{3}}}{4})$

It simplifies a bit further, but i'd really like confirmation that I have interpreted this question correctly.

Thanks.

 

[pasmith]

What you have here is a definite integral: the integral of $(3 + 4t)^{5/3}$ between $x_0$ and $x$. It's no different from if you were asked to evaluate $\int_a^b (3 + 4t)^{5/3}\,dt$.

When writing definite integrals it is bad practice to use the same symbol as both a limit of the integral and as the dummy variable. And when a question uses a particular symbol, you should use that same symbol in your answer.

 

[HallsofIvy]

Yes, when you change the variable, you have to change the limits of integration as well.

Your integral has $x_0$ and $x$ as limits of integration and you let u= 3+ 4t. At the lower limit, $t= x_0$ so $u= 3+ 4x_0$. At the upper limit, $t= x$ so $u= 3+ 4x$. Your integral should be
$$\frac{1}{4}\int_{3+4x_0}^{3+ 4x} u^{5/3} du$$

You final result, whether you integrate with respect to t or u, should be a function of x.

 

[BOAS]

Yes, when you change the variable, you have to change the limits of integration as well.

Your integral has $x_0$ and $x$ as limits of integration and you let u= 3+ 4t. At the lower limit, $t= x_0$ so $u= 3+ 4x_0$. At the upper limit, $t= x$ so $u= 3+ 4x$. Your integral should be
$$\frac{1}{4}\int_{3+4x_0}^{3+ 4x} u^{5/3} du$$

You final result, whether you integrate with respect to t or u, should be a function of x.

Hello,

does this mean that here $I = (\frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4t_{0})^{\frac{8}{3}}}{4})$, where there is a t, I need to substitute in $(3+ 4x)$ and $(3+4x_{0})$

To get $I = (\frac{3}{5} \frac{(3+4(3+4x))^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4(3 + 4x_{0})^{\frac{8}{3}}}{4})$?

 

[BOAS]

What you have here is a definite integral: the integral of $(3 + 4t)^{5/3}$ between $x_0$ and $x$. It's no different from if you were asked to evaluate $\int_a^b (3 + 4t)^{5/3}\,dt$.

When writing definite integrals it is bad practice to use the same symbol as both a limit of the integral and as the dummy variable. And when a question uses a particular symbol, you should use that same symbol in your answer.

Hi,

now that you've said it, I do recall my lecturer discussing dummy variables.

Thanks.

 

[BvU]

Hello,

does this mean that here $I = (\frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4t_{0})^{\frac{8}{3}}}{4})$, where there is a t, I need to substitute in $(3+ 4x)$ and $(3+4x_{0})$

To get $I = (\frac{3}{5} \frac{(3+4(3+4x))^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4(3 + 4x_{0})^{\frac{8}{3}}}{4})$?

No ! Now you've done the transform twice.

You work out an expression for $\frac{1}{4}\int_a^b u^{5/3} du$ and fill in a = 3 + 4x₀ and b = 3 + 4x

 

[BOAS]

No ! Now you've done the transform twice.

You work out an expression for $\frac{1}{4}\int_a^b u^{5/3} du$ and fill in a = 3 + 4x₀ and b = 3 + 4x

Thanks for clarifying that, I suspected this would be the case.