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I'm trying to solve an integral and I don't remember how to when a variable is in the integral symbol, if that makes sense. I'm not sure what the correct terminology is.

So day I have an integral like:

int

How do I solve it? What do I have to do differently?

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    @Arturo: I wonder the same thing... too bad we couldn't get a clarification.2011-07-13

3 Answers 3

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When you are computing the definite integral $\int_a^b f(t)dt$ your usual procedure has been:

(i) Find an antiderivative $F(t)$ of $f(t)$. That means, find a function $F(t)$ such that F'(t)=f(t).

(ii) Then your "answer" is $F(b)-F(a)$.

The procedure is exactly the same if $a$ and $b$ are not constants, but are functions of some variable $x$. So (ii) remains unchanged.

In your particular integral, we have $a=a(x)=x^3$ and $b=b(x)=-3$. The "lower" limit of integration is a constant, which you can think of as a constant function, and the upper limit is the function $x^3$.

Let's calculate. We have $f(t)=t^2$. So one antiderivative of $f(t)$ is $F(t)$, where $F(t)=\frac{t^3}{3}$

Now do the familiar substitution process. We get $\int_{-3}^{x^3}t^2dt=F(x^3)-F(-3)=\frac{(x^3)^3 -(-3)^3}{3}$

We might want to simplify this to $\frac{x^9+27}{3}$

Comments: The variable $t$ is called a "dummy variable" roughly because ultimately it plays no role in the answer. You would get exactly the same thing if you had $u$ as the variable, or $w$, instead of $t$.

In principle, you could also have used the letter $x$ as the dummy variable. But don't ever do it when one or both of the limits of integration involves $x$. Although it is technically not wrong, the chances of getting confused are too high.

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To be explicit, you can do this $f(x) = \int_{-3}^{x^3} t^2\, dt = {t^3\over 3}\bigg|_{-3}^{x^3} = {x^9 + 27\over 3}.$ There is nothing scary here; just treat $x$ as if it were a constant and you will be fine.

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Easily done; let $u=x^3$, proceed with the integration as usual, and then replace the $u$ in your result with $x^3$.

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    @In between: you are integrating wrong; the integral of $t^2$ is not $t^2$, it's $\frac{1}{3}t^3$ (plus a constant, but for definite integral evaluation you don't need the constant).2011-06-15