I have a question like
Find a lower bound and an upper bound for the area under the curve by finding the minimum and maximum values of the integrand on the given integral:
$$ \int_1^6t^2-6t+11 \ dt $$
It asks for two answers; a minimum area and a maximum area.
So, I integrate this;
$$ \left(\frac{t^3}{3}-3t^2+11t\right)\Bigg|_1^6 $$
I know I have a minimum at $x = 3$ because;
$$ f(t) = t^2-6t+11 \\ f'(t) = 2t-6 = 0 \\ 2(t-3) = 0 \\ t = 3 \\ f(5) = 4 \\ f(1) = -4 \\ $$
Very confused by what is going on when it asks for a maximum area and a minimum area.
$$f(t)=t^2-6t+11=(t-3)^2+2$$
which is a convex quadratic function. It has a unique minimum.
The boundary points are $1$ and $6$, we can evaluate these points and conclude that $f(6)$ is bigger. $f(6)$ is the global maximum. (or just observe that the quadratic curve is symmetrical about $t=3$ and the value increases the further away we are from $3$).
Hence, $$\forall t \in [1,6], f(3) \leq f(t) \leq f(6)$$
$$\int_1^6 f(3) dt \leq \int_1^6 f(t) dt \leq \int_1^6 f(6) dt$$
$$5 f(3) \leq \int_1^6f(t)dt \leq 5f(6)$$
Of course, $f(3)$ and $f(6)$ can be evaluated easily.