# Conceptual doubt in Definite Integration regarding limits of $\int_{0}^{2}x\cdot\sqrt{x+2}\ dx$

Question:

Evaluate $$I=\int_{0}^{2}x\cdot\sqrt{x+2}\ dx$$.

My Approach:

Put $$x+2=t^{2}$$. Now changing limits: Lower Limit $$\left(x=0\right)\ :\ t^{2}=0+2=2$$ $$\to$$ $$t=\pm\sqrt{2}$$

How do I know if i have to take the t with + or - sign? Same confusion is with the upper limit. Sorry if this might be too bad of a question to ask but I have only began with definite integration. I know This doubt is really specific because taking some other substitution would not have landed me in this situation. But how do I proceed this way?

Edit:

I think everyone is right, It is just a matter of preference. What alex is saying is:

Let me take $$-\sqrt{2}$$ as the lower limit and $$-2$$ as the upper limit, since I have substituted $$t^{2}=x+2$$ We have, $$dx=2tdt$$ and the integral becomes: $$I=2\int_{-\sqrt{2}}^{-2}\left|t\right|\cdot t\cdot\left(t^{2}-2\right)dt$$ which gives the same answer. Here we will open |t| with - sign because we know it is varying in negative number.

We can even take lower limit as $$-\sqrt{2}$$ and upper limit as $$+2$$ and evaluate the integral which gives the same thing but this time I have to split the integral because of the mod, specifically: $$I=2\int_{-\sqrt{2}}^{2}\left|t\right|\cdot t\cdot\left(t^{2}-2\right)dt$$

$$I=2\int_{-\sqrt{2}}^{0}\left(-t\right)\cdot t\cdot\left(t^{2}-2\right)dt+2\int_{0}^{2}\left(+t\right)\cdot t\cdot\left(t^{2}-2\right)dt$$ .

Both give the same answer. Obviously integrating from $$+\sqrt{2}\to2$$ is easy but it's nice to get the concepts right.

• You have upvoted a bad answer. Your mistake is in writing $x+2=t^{2}$ instead of the direct substitution $t =\sqrt {x+2}$ the positive square root of $x+2$ which allows you to evlauate the given integral. Sep 2, 2021 at 23:17
• @KaviRamaMurthy Agreed. $t=\sqrt{x+2}$ is basically the same thing but helps avoid the above confusion and we can still square this and get $x=t^{2}-2$ and evaluate the integral. Sep 3, 2021 at 2:59
• You can also just use the property - $\int_0^a f(x)dx = \int_0^a f(a-x)dx$...it'll make your life a lot easier Oct 2, 2021 at 6:56

Instead of starting with the equation $$t^{2}=x+2$$ you should start with $$t =\sqrt {x+2}$$ the positive square root of $$x+2$$.

(The notation $$\sqrt {x+2}$$ in the question stands for the positive square root of $$x+2$$). So you take $$t >0$$. The intergal w.r.t. $$t$$ is from $$\sqrt 2$$ to $$2$$.

• Okay so since $\sqrt{x+2}=t$, I have to consider the positive value of t? Sep 2, 2021 at 11:57
• Yes, integrate w.r.t. $t$ from $\sqrt 2$ to $2$. @Vega Sep 2, 2021 at 11:58
• OP made the substitution $t^2=x+2\implies t=\pm\sqrt{x+2}$ and not just $+\sqrt{x+2}$. Alex's answer is more convincing in this regard Sep 2, 2021 at 13:07

In my humble opinion the correct answer is: it doesn't matter, since $$\sqrt{t^2}=|t|$$, so if you choose $$\sqrt 2$$ you have to choose $$|t|=t$$, if you choose $$-\sqrt 2$$ you have to split the cases, namely $$|t|=-t$$ between $$-\sqrt 2$$ and $$0$$, and $$|t|=t$$ between $$0$$ and $$2$$. I admit of course that the choice $$\sqrt 2$$ is way more rational than the other one, but both are correct.

• It is accepted by all Mathematicians that $\sqrt x$ stands for the positive square root of $x$ for a positive real number $x$ . Sep 2, 2021 at 13:14
• @Kavi Rama Murthy Yes, and $|t|$ is indeed the positive square root of the positive real number $t^2$.
– alex
Sep 2, 2021 at 13:16
• You are not asked to find all substitutions that give the value of the integral. You take t to be the positive square root of x+2 and that gives you the value of the integral. There is no need to ever bring in negative values of t and use |t| anywhere. It is strange that so many users upvoted your misleading answer. @vega Sep 3, 2021 at 0:20
• You should delete the first sentence in your answer. You are trying to say that my answer is not correct. Is this really being humble? Sep 3, 2021 at 5:07

I want to add a hopefully last perspective on this. You can in fact choose any combination of the limits you want: With $$f(t) =(t^2-2)\cdot |t|\cdot 2t$$ we have $$\int_{\sqrt 2}^2f(t)\,dt = \int_{-\sqrt 2}^2f(t)\,dt = \int_{\sqrt 2}^{-2}f(t)\,dt = \int_{-\sqrt 2}^{-2}f(t)\,dt.$$ This is because all of them are equal to $$\int_0^2x\sqrt{x+2}\,dx$$ through the substitution $$x:=t^2-2$$. The first and last version correspond to the standard substitutions $$t=\pm\sqrt{x+2}$$. But all of them work if we read the theorem of substitution in reverse, which amounts to letting $$x$$ depend on $$t$$ and not the other way round.

Let's look at it formally. The precise statement of substitution is, if $$\varphi$$ is $$C^1$$ and $$f$$ is continuous on the relevant regions, then $$\int_{\varphi(a)}^{\varphi(b)}f(x)\, dx = \int_a^b f(\varphi(t))\varphi'(t)\,dt.$$ I've written it in the unusual order, because we want to use the above from left to right. Then with $$f(x) = x\sqrt{x+2}$$, $$\varphi(t)=t^2-2$$, $$\varphi(a)=0$$ and $$\varphi(b)=2$$, we get the result. Note that $$a$$ and $$b$$ are any numbers that satisfy $$\varphi(a)=0$$ and $$\varphi(b)=2$$. We see that we always have complete freedom of $$a$$ and $$b$$ in cases like this.

Both of the substitutions $$t=\sqrt{x+2}$$ and $$t=-\sqrt{x-2}$$ are valid transformations of the integral, as the functions $$x\mapsto\sqrt{x+2}$$ and $$x\mapsto-\sqrt{x+2}$$ are both invertible on the interval $$[0,2]$$. This condition is necessary, as the term $$\left(\sqrt{x+2}\right)'=\frac{1}{2\sqrt{x+2}}$$ does not appear in the integrand.

For further discussion of the theory behind integration by substitution, see Michael Spivak's Calculus, in particular, chapter 19: integration in elementary terms.