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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
    Commented Jun 21, 2017 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
    Commented Jun 21, 2017 at 6:53
  • $\begingroup$ This is exactly how it is phrased in the book $\endgroup$
    – Computer
    Commented Jun 21, 2017 at 7:25
  • $\begingroup$ @Macavity, it says 'upper bound' and 'lower bound' which makes sense. I will edit the title accordingly $\endgroup$
    – Yuriy S
    Commented Jun 21, 2017 at 9:53
  • $\begingroup$ But the bounds are already included $[1,6]$ $\endgroup$
    – Computer
    Commented Jun 21, 2017 at 10:20

1 Answer 1

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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
    Commented Jun 21, 2017 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? $\endgroup$ Commented Jun 21, 2017 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
    Commented Jun 21, 2017 at 22:42
  • $\begingroup$ Also where does 5 come from? $\endgroup$
    – Computer
    Commented Jun 21, 2017 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. $\endgroup$ Commented Jun 21, 2017 at 22:44

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