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I'm trying to solve this question using the definition of integral:

$$\int^5_2 (4-2x)dx$$

Definition of integral:

We define first the inferior and superior sum:

Let $f:[a,b]\to \mathbb R$ be a bounded function and $P=\{t_0,\ldots,t_n\}$ a partition of [a,b], so we define:

$$s(f;P)=\sum_{i=1}^nm_i(t_i-t_{i-1})$$

$$S(f;P)=\sum_{i=1}^nM_i(t_i-t_{i-1})$$

Where $m_i=\inf\{f(x);x\in [t_i,t_{i-1}]\}$ and $M_i=\sup\{f(x);x\in [t_i,t_{i-1}]\}$ and ().

No we define the inferior and superior integral:

Inferior integral: $\underline\int^b_a=\sup_P s(f;P)$

Superior integral: $\bar\int^b_a=\inf_P s(f;P)$

At last, we define the integral being the superior or inferior integral when they are the same.

My attempt

I know:

$s(f,P)=\sum_{i=1}^n(4-2t_i)(t_i-t_{i-1})$ (because the function is decreasing)

But I don't know how to take the supreme of this sum.

Thanks.

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    $\begingroup$ "Solve" is the wrong word. One may evaluate an expression (such as an integral). One solves an equation or solves a problem. $\endgroup$
    – Michael Hardy
    Jul 21, 2014 at 12:17
  • $\begingroup$ Thank you for the remark :) $\endgroup$
    – user42912
    Jul 21, 2014 at 14:07

4 Answers 4

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The contributions from fellow users Mhenni Benghorbal and oks pave the way.

Using equidistant spacings one can perform the summation like this: $$ \begin{align} s(f,P') &= \sum_{i=1}^n(4-2t_i)(t_i-t_{i-1}) \\ &= \left(\sum_{i=1}^n 4 - 2 \left(a + \frac{b-a}{n} i \right) \right)\frac{b-a}{n} \\ &= \left(4n - 2 \left(a n + (b-a) \frac{n+1}{2} \right) \right) \frac{b-a}{n} \\ &= \left(4 - 2a - (b-a)\left(1+ \frac{1}{n}\right) \right) (b-a) \\ &= \left(4 - 2a - (b-a)\right) (b-a) - \frac{(b-a)^2}{n} \\ \Rightarrow \sup s(f, P') &= \left(4 - 2a - (b-a) \right) (b-a) \\ &=4(b-a) -(b+a)(b-a) \\ &=4(b-a) -(b^2 - a^2) \end{align} $$ and $$ \begin{align} S(f,P') &= \sum_{i=1}^n(4-2t_{i-1})(t_i-t_{i-1}) \\ &= \left(\sum_{i=1}^n 4 - 2 \left(a + \frac{b-a}{n} (i-1) \right) \right)\frac{b-a}{n} \\ &= \left(4n - 2 \left(a n + (b-a) \frac{n-1}{2} \right) \right) \frac{b-a}{n} \\ &= \left(4 - 2a - (b-a)\left(1-\frac{1}{n}\right) \right) (b-a) \\ &= \left(4 - 2a - (b-a)\right) (b-a) + \frac{(b-a)^2}{n} \\ \Rightarrow \inf S(f,P') &= \left(4 - 2a - (b-a) \right) (b-a) \\ &=4(b-a) -(b^2 - a^2) \end{align} $$

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  • $\begingroup$ So isn't possible to evaluate this integral without using equidistant spacings? $\endgroup$
    – user42912
    Jul 22, 2014 at 9:50
  • $\begingroup$ I would not go that far. The above works, because the $t_i$ were expressed in terms of $a$, $b$ and $n$ . This is possible for non-equidistant spacings as well. Maybe it makes integration easier for some functions. $\endgroup$
    – mvw
    Jul 22, 2014 at 12:48
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Hint: Since you know that $s(f,P)<S(f,P')$ for any two divisions $P,P'$, it is enough to find, for each $\epsilon$, such a division $P$ that $$-9-\epsilon < s(f,P)< S(f,P) < -9+\epsilon$$

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It is always much easier to use the Mean Value Theorem on the function $g(x) = 4x - x^2$ ni every interval on the partition. That is for each $i$ there exists $c_i \in [t_{i- 1}, t_i] $ such that

$$ \dfrac{g(t_i) - g(t_{i-1})}{ t_i - t_{i- 1} } = g'(c_i) = (4 - 2x)|_{x = c_i} = 4 - 2c_i = f(c_i)$$

Now try to use the fact that $m_i \ge f(c_i) \ge M_i$ since $ t_{i-1} \le c_i \le t_i $ and the function $f$ is increasing.

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  • $\begingroup$ Yes, but isn't it possible to solve this integral using only the definition of integral as I was trying to do? $\endgroup$
    – user42912
    Jul 21, 2014 at 10:39
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Hint: if you are integrating over an interval $[a,b]$ then

$$ \Delta t_i = t_i-t_{i-1} = \frac{b-a}{n}. $$

Also, your function is a decreasing function then you should know what $m_i$ and $M_i$ are.

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  • $\begingroup$ He does know what $m_i$ and $M_i$ are. Also, you only supplied one possible partition $P$. OP must find supremum over all possible partitions. $\endgroup$
    – 5xum
    Jul 21, 2014 at 10:24
  • $\begingroup$ My definition is different, the delta in my definition is not fixed. $\endgroup$
    – user42912
    Jul 21, 2014 at 10:27
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    $\begingroup$ For every partition P there is a partition P' where P' has fixed delta equal to the smallest $t_i - t_{i-1}$ in P. Since the function is decreasing, s(f; P) < s(f; P'), and so $$\sup_P s(f;P) \le \sup_{P'} s(f;P')$$ But P' is just a (specialised) partition, so $$\sup_{P'} s(f;P) \le \sup_{P} s(f;P)$$ So $$\sup_P s(f;P) = \sup_{P'} s(f;P')$$ and from there you can use Mhenni Benghorbal $\endgroup$
    – oks
    Jul 21, 2014 at 11:14
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    $\begingroup$ @oks Sounds good. Finally using an equidistant partition allows to perform the summation. $\endgroup$
    – mvw
    Jul 21, 2014 at 11:20

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