If a function f is defined by f(x) = S from 0 to x (1 / (1 + t^4)) dt, is f(1) = to 1/2?
I take it I have to integrate that thing, how do I do it?
we can say no without actually evaluating the integral.
$\displaystyle f(1)=\int_{0}^{1}\frac{1}{1+t^4}dt > \frac{1}{2}$
By camparison the minimum of $\displaystyle \frac{1}{1+t^4}$ on [0,1] is 1/2 and the max is 1 so
$\displaystyle \int_{0}^{1}\frac{1}{2}dt < \int_{0}^{1}\frac{1}{1+t^4}dt < \int_{0}^{1}1dt$
equallity could only happen if the integrand was constant on [0,1]
I hope this helps.
Good luck.
let
$\displaystyle h(t)=\frac{1}{1+t^4}$
$\displaystyle h(0)=\frac{1}{1+0^4}=1$
$\displaystyle h(1)=\frac{1}{1+1^4}=\frac{1}{2}$
If your really want to be rigorous you could verify this with the first and second derivative test, but the function is postitive and decreasing for all t > 0 so its max is at the left end point and its min is at the right end point.
I hope this clears it up.