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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.

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  • $\begingroup$ A function can have a minimum and maximum value where the derivative isn't $0$. Specifically, where its derivative isn't defined and, perhaps more relevant to this problem, at the end points of where the function is defined. Of course, just as for when the derivative is $0$, you have to actually check whether these are max / min values. $\endgroup$ – Arthur Jun 21 '17 at 6:47
  • $\begingroup$ Are you sure you have reproduced the question correctly or is something lost in paraphrasing? As stated, the question seems a bit absurd - the definite integral is a rational number, so speaking of its maximum and minimum doesn't make much sense. $\endgroup$ – Macavity Jun 21 '17 at 6:53
  • $\begingroup$ This is exactly how it is phrased in the book $\endgroup$ – Computer Jun 21 '17 at 7:25
  • $\begingroup$ @Macavity, it says 'upper bound' and 'lower bound' which makes sense. I will edit the title accordingly $\endgroup$ – Yuriy S Jun 21 '17 at 9:53
  • $\begingroup$ But the bounds are already included $[1,6]$ $\endgroup$ – Computer Jun 21 '17 at 10:20
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$$f(t)=t^2-6t+11=(t-3)^2+2$$

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.

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  • $\begingroup$ I do not understand this. $\endgroup$ – Computer Jun 21 '17 at 22:31
  • $\begingroup$ which part? do you understand why $f(3) \leq f(t) \leq f(6)$? or is it other part that you are confused about? $\endgroup$ – Siong Thye Goh Jun 21 '17 at 22:39
  • $\begingroup$ Is $f(3)$ chosen because the zero of $(t-3)^2+2 = 3$? And if so, why choose to include $f(6)$ and not $f(1)$? $\endgroup$ – Computer Jun 21 '17 at 22:42
  • $\begingroup$ Also where does 5 come from? $\endgroup$ – Computer Jun 21 '17 at 22:44
  • $\begingroup$ $f(3)$ is chosen because $(t-3)^2+2 \geq 2$ and this minimal is attaiend when $t=3$, that is when we zero out the first term. We can evaluate $f(6)$ and $f(1)$ and see which number is bigger. $\endgroup$ – Siong Thye Goh Jun 21 '17 at 22:44

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