why cannot the limits be $-1$ and $-2$ I came across a problem in definite integral as  : Evaluate
$$I=\int_{0}^{3} x\sqrt{1+x}\:dx$$
By the substitution $1+x=t^2$
so book has given lower and upper limits as $t=1$ and $t=2$ which is obtained as
$t^2=1$ $\implies$ $t=1$ and
$t^2=4$ $\implies$ $t=2$
we get $$I=\int_{1}^{2} (t^2-1)(t)(2t)dt=2\int_{1}^{2}(t^4-t^2)dt=\frac{116}{15}$$
but why cant we take limits as $t=-1$ and $t=-2$ then we get
$$I=2\int_{-1}^{-2}(t^4-t^2)dt$$
since $t^4-t^2$ is even function we have $$I=\frac{-116}{15}$$
what is the mistake in my analysis?
 A: The book does a bad service to students, in my opinion. The substitution should be $t=\sqrt{1+x}$, making it clear that $t\ge0$ and also providing for automatic substitution of the bounds.
Of course you can also do the substitution $t=-\sqrt{1+x}$, so the bounds are $-1$ and $-2$, but the integral becomes
$$
\int_{-1}^{-2}-(t^2-1)t\cdot 2t\,dt
$$
so the final result is the same as with $t=\sqrt{1+x}$.
The problem is that the function $t\mapsto t^2-1$ is not injective, so you have to choose a branch where it is; for instance $t\ge0$ or $t\le0$. The substitution spelled out as $t=\sqrt{1+x}$ or
$$
\begin{cases}
x+1=t^2\\
t\ge0
\end{cases}
$$
avoids the problem.
A: I think you can solve it without changing variable.
$$I=\int_{0}^{3} x\sqrt{1+x}\:dx=\int_{1}^{4} (x-1)\sqrt{x}\:dx=\int_{1}^{4} (x\sqrt{x}-\sqrt{x})\:dx=\frac{2}{5}x^\frac{5}{2}-\frac{2}{3}x^\frac{3}{2}\mathcal{|_{1}^{4}}=\frac{116}{15}$$
A: Because in the substitution, $ t = \sqrt{x+1}$ and $\sqrt{x+1}\geq 0$
$ t$ should be bigger than or equal to  $0$.
Or when $ t^2 = x+1$ the $ \sqrt{x+1}= |t|$.
