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I tried to find the integral of the function

$$\int_0^2x(x+2)^\frac{1}{2}dx$$
I substituted $$x+2=t^2$$ but while converting limits I found $t=2,-2$ as upper limit. When integrating with these different limits I get different answers while the correct is with only positive one? Why can't the one with negative one be correct?
If I do the same for the lower limit, the same situation arises.

coldnumber
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Sikander
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    The substitution is either $x+2=t^2$ where $t\ge 0$, or where $t\le 0$. We are really setting $t=\sqrt{x+2}$ or $t=-\sqrt{x+2}$. Of course I prefer plus, but whichever you want to use is OK as long as you are consistent. – André Nicolas Jul 16 '15 at 02:30
  • That you get the same answer either way is what I showed in my answer posted below. ${}\qquad{}$ – Michael Hardy Jul 16 '15 at 02:51

2 Answers2

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$$ \int_0^2x(x+2)^{1/2}dx $$ If $t^2 = x+2$ then $2t\,dt = dx$ and $t^2 - 2 = x$.

As $x$ goes from $0$ to $2$, then $t^2$ goes from $2$ to $4$, so $t$ goes from${}\,\ldots\,{}$ what, to what?

We could let $t$ go from $\sqrt 2$ to $2$. Then we have $$ \int_{\sqrt 2}^2 (t^2-2) t\,(2t\,dt) = 2 \int_{\sqrt 2}^2 (t^4 - 2t^2)\,dt = 2\left[ \frac{t^5} 5 - \frac{2t^3} 3 \right]_{\sqrt 2}^2 = \frac {112 + 16\sqrt 2} {15}. $$

Or we could let $t$ go from $-\sqrt 2$ to $-2$. Then we have $$ \int_{-\sqrt 2}^{-2} (t^2-2)(-t)(2t\,dt) = -2\int_{-\sqrt 2}^{-2} (t^4-2t^2)\,dt = - 2 \left[ \frac{t^5} 5 - \frac{2t^3} 3 \right]_{-\sqrt 2}^{-2} = \frac{112 + 16\sqrt 2}{15}. $$

The answer is the same either way.

Michael Hardy
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  • Why did you take -t.What it mean to me is that t is negative here and you tried to change sign and made the situation same as was with the positive limits. – Sikander Jul 16 '15 at 11:05
  • @Raul : If $t^2=x+2$ then $t=\pm\sqrt{x+2}$. If that goes from $-\sqrt2$ to $-2$, then $t=-\sqrt{x+2}$, so $\sqrt{x+2}$ is $-t$. ${}\qquad{}$ – Michael Hardy Jul 16 '15 at 17:56
  • Dont you think underroot of x+2 can itself take a negative number.It is looking as you are taking underroot of x+2 as modulus – Sikander Jul 16 '15 at 18:51
  • @Raul : $x+2$ would be positive but $t$ would be negative. ${}\qquad{}$ – Michael Hardy Jul 16 '15 at 20:05
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Let $\simeq$ mean "to be equal to up to a constant". Then we have $$ \int x(x+2)^{1/2} dx = \int \bigg[ (x+2)^{3/2} - 2(x+2)^{1/2} \bigg] d(x+2) \simeq \frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2}. $$ Can you complete?

Yes
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