Integration - Change of Variable

1. Nov 12, 2014

BOAS

1. The problem statement, all variables and given/known data

Use integration by substitution to evaluate the integral,

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

2. Relevant equations

3. 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.

2. Nov 12, 2014

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.

3. Nov 12, 2014

HallsofIvy

Staff Emeritus
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.

4. Nov 12, 2014

BOAS

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

5. Nov 12, 2014

BOAS

Hi,

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

Thanks.

6. Nov 14, 2014

BvU

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

7. Nov 14, 2014

BOAS

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