# Integral with undefined constants

I have no idea how to do this problem at all.

$$\int_{0}^{a} x \sqrt{x^2 + a^2} dx$$

I set $u$

$u = x^2 + a^2$

$du = 2x dx$

$dx=du/2x$

$$\int_{0}^{a} x \sqrt{u}*du/2x = \frac {1}{2} \int_{0}^{a} u^{\frac{1}{2}} dx$$

$\frac{1}{2} * \frac{2}{3} u^{\frac{3}{2}}$

From here I am stuck.

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When you change the variable from $x$ to $u=x^2+a^2$ (typo alert!), you need to change the interval of integration as well. When $x=0$, how much is $u$? When $x=a$, how much is $u$? The interval changes, so the evaluation changes. But once you get there, you just plug in the values, and leave the answer in terms of $a$. – Arturo Magidin Apr 27 '12 at 21:15
I have never heard of that, and for some reason I keep getting the correct answers. How do I know how to change the intervals? – user138246 Apr 27 '12 at 21:19
You can just plug $x^2+a^2$ back in where $u$ is, and then evaluate that function at $a$, and subtract that function evaluated at $0$. – NeuroFuzzy Apr 27 '12 at 21:20
@Jordan: That's definitely in your book. For indefinite integrals it doesn't matter. For definite integrals, you either need to go back to the original variable, or you need to change the intervals. otherwise, you do not get the right answers. For example, $\int_0^19x^2\,dx = 3x^3|_0^1 = 3$. If you do the change $u=3x$, and forget to do the change of intervals, you get $$\int_0^19x^2\,dx"="\frac{1}{3}\int_0^1u^2\,du = \frac{1}{3}\left(\frac{1}{3}u^3|_0^1\right) = \frac{1}{3}\left(\frac{1}{3}\right) = \frac{1}{9}.$$ – Arturo Magidin Apr 27 '12 at 21:24
I have seen your question"I can't do math". Maybe this one partly shows this problem.I think you need to turn to your teacher for a help.As for me,there're three methods to solve problems in calculus:derivative,integrate, and series.I regard them as tools because they are not just used to solve some specific problems(kinds of problems you meet them before ). – 89085731 Apr 28 '12 at 1:23

It's mostly fine. However:

First. When you do a change of variable/substitution in a definite integral, you need to change the interval of integration.

Here you are setting $u=x^2+a^2$. That means that when $x=0$, you have $u=0^2+a^2 = a^2$; and when $x=a$, you get $u=a^2+a^2=2a^2$. So the integral will not be an integral from $0$ to $a$ when it is an integral with respect to $u$, it will be an integral from $a^2$ to $2a^2$. So you go from $$\int_0^a x\sqrt{x^2+a^2}\,dx$$ to $$\frac{1}{2}\int_{a^2}^{2a^2}u^{1/2}\,du.$$

Second. You just plug in: \begin{align*} \int_0^a x\sqrt{x^2+a^2}\,dx &= \frac{1}{2}\int_{a^2}^{2a^2}u^{1/2}\,du &\text{(through all the work you've done)}\\ &= \frac{1}{2}\left.\left(\frac{2}{3}u^{3/2}\right|_{a^2}^{2a^2}\right)\\ &= \frac{1}{2}\left(\frac{2}{3}(2a^2)^{3/2} - \frac{2}{3}(a^2)^{3/2}\right)\\ &= \frac{1}{2}\frac{2}{3}2^{3/2}a^{2(3/2)} - \frac{1}{2}\frac{2}{3}a^{2(3/2)}\\ &= \frac{2^{3/2}}{3}a^3 - \frac{1}{3}a^{3}\\ &= \left(\frac{2^{3/2}}{3} - \frac{1}{3}\right)a^3\\ &= \left(\frac{2\sqrt{2}-1}{3}\right)a^3. \end{align*} That's it: since the problem is in terms of $a$, the answer will also be in terms of $a$. (I used the fact that $2^{3/2} = (2^{1/2})^3 = (\sqrt{2})^3 = (\sqrt{2})^2\sqrt{2} = 2\sqrt{2}$).

The alternative way is to do the indefinite integral first, and then use the antiderivative: to compute $$\int_0^ax\sqrt{x^2+a^2}\,dx$$ we first do the indefinite integral (using your substitution): \begin{align*} \int x\sqrt{x^2+a^2}\,dx &= \frac{1}{2}\int u^{1/2}\,du &\text{(setting }u=x^2+a^2\text{)}\\ &= \frac{1}{2}\left(\frac{2}{3}u^{3/2}\right) + C\\ &= \frac{1}{3}u^{3/2}+C\\ &= \frac{1}{3}(x^2+a^2)^{3/2} + C. &\text{(going back to }x\text{)} \end{align*} Having done that, we can use this antiderivative (with any particular value for $C$) to obtain the original definite integral, via the Fundamental Theorem of Calculus: \begin{align*} \int_0^a x\sqrt{x^2+a^2}\,dx &= \left.\left(\frac{1}{3}(x^2+a^2)^{3/2}\right|_0^a\right)\\ &= \left(\frac{1}{3}(a^2+a^2)^{3/2}\right) - \left(\frac{1}{3}(0^2+a^2)^{3/2}\right)\\ &= \frac{1}{3}(2a^2)^{3/2} - \frac{1}{3}(a^2)^{3/2}\\ &= \frac{2^{3/2}}{3}a^3 - \frac{1}{3}a^3\\ &= \frac{2\sqrt{2}-1}{3}a^3, \end{align*} same answer as above.

What you should not do is do a substitution in a definite integral and not correct the limits of integration. The way I tell my students this is to remember the following maxim:

If you change the variable, you have to change the limits.

You should not rely on leaving the limits as they were thinking "I'll just go back to $x$ when I'm done with the integral." That's just asking for trouble (like, forgetting to go back to $x$), and it may be difficult to unwind if you have to make several substitutions along the way (or use other methods of integration). Better to do the simple computations needed to change the interval (which amount to just plugging the limits into the substitution formula) and then not have to worry about it.

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This is incredibly difficult to image graphically, I assumed a subsitution rule was just a different want to write the problem but I am physically changing the way the graph looks so how am I not changing the area under the graph? – user138246 Apr 27 '12 at 21:26
@Jordan: Say you do the substitution $u=2x$. In order to cover the same are, if your original $x$ took values between, say, $0$ and $1$, then $u$ should take values between $0$ and $2$ in order for $u$ to cover the same territory as $x$ did. ($u$ covers the region "twice as fast" as $x$, so in order to finish at the same time as $x$ did, it has to run "twice as far", so instead of just going from $0$ to $1$, it needs to go from $0$ to $2$). – Arturo Magidin Apr 27 '12 at 21:29

You want to find

$$\int\limits_0^a {\sqrt {{x^2} + {a^2}} xdx} =F(a)$$

With a change of variables, namely $u=x^2+a^2$, we get that $du=2xdx$, from where the integral is

$$\int\limits_0^a {\sqrt {{x^2} + {a^2}} xdx} = \frac{1}{2}\int\limits_{a^2}^{2{a^2}} {\sqrt u du}$$

Note that I changed $a$ to $2a^2$, since if $x=a$, then $$u = {a^2} + {a^2} = 2{a^2}$$

and similarily, for $x=0$, $u=a^2$. Now we simply use $\mathcal{FTC}$ to get

$$\int\limits_0^a {\sqrt {{x^2} + {a^2}} xdx} = \frac{1}{2}\int\limits_{{a^2}}^{2{a^2}} {\sqrt u du} = \frac{1}{3}\left[ {2\sqrt 2 {a^3} - {a^3}} \right]$$

since a primitive of $\sqrt u$ is $\frac{2}{3}{u^{3/2}}$. As a consequence, your problem is solved,

$$\int\limits_0^a {\sqrt {{x^2} + {a^2}} xdx} = \frac{{2\sqrt 2 - 1}}{3}{a^3}$$

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You shouldn't call your new variable $x$, that's confusing because your old variable is called $x$. So let's call the new one $u$ and set $u = x^2 + a^2$. Now you want to compute $du$. You do this by taking the derivative of $u$: $u^\prime = \frac{du}{dx} = 2x$. Note that since $a$ is a number, its derivative is zero. So we have $du = 2x dx$.

As for the range of the integral, from $x=0$ to $x=a$, we need to adapt this for $u$: If $x=0$ we get $u = a^2$ and if $x=a$ we get $u=2a^2$. (by substituting $0$ and $a$ into $u=x^2 + a^2$)

Substituting into the integral we get $$\int_{x=0}^{x=a} x \sqrt{x^2 + a^2} dx = \int_{u=a^2}^{u=2a^2} x \sqrt{u} \frac{du}{2x} =\frac{1}{2} \int_{u=a^2}^{u=2a^2} \sqrt{u} du$$

I'll let you compute the integral of $\sqrt{u}$, it's not hard.

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I think the $x$ in $x=x^2+a^2$ was just a typo; note that the paragraph before that he says "I set $u$", and the line after he computes $du$.... I corrected it to that. – Arturo Magidin Apr 27 '12 at 21:33
@ArturoMagidin Yes, I think so too. But I think he's confused so I decided to elaborate on his typo. – Rudy the Reindeer Apr 27 '12 at 21:34