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?

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|$.
