The question:
$\displaystyle \int_{-1}^{0} \sqrt{t^2 + t^4} dt$
I'm stumped. Not sure how to attempt this.
Any assistance would be great.
How about first taking $\displaystyle t^2$ from within the square root, and move it outside as $\displaystyle |t|$:
$\displaystyle \sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$
where $\displaystyle |t|=-t$, because we find ourselves in the range $\displaystyle -1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...
$\displaystyle \displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$
It is quite obvious that the second term of the 'identity' (1) is an odd function ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.
In fact I should have writtent $\displaystyle \sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$ , but in this case is irrelevant since the
variable t is already positive in the new integration's limits.
Tonio
Let $\displaystyle -u = t$, then $\displaystyle dt = -du$, then:
$\displaystyle \displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}. $
Now we can write $\displaystyle \sqrt{t^2+t^4} = t\sqrt{1+t^2}$, as our limits are non-negative.