1
$\begingroup$

Using the upper and lower sums, I've tried to solve $\int_1^e\frac{1}{x}dx$ in the following way.

Let $$\Delta x = \frac{\mathit e -1}{n}$$ and let a partition $P$ be given by $$ P = \{1,1+\Delta x, 1+2\Delta x,...,1+(n-1)\Delta x,\mathit e\}$$

Since $1/x$ is a monotonically decreasing function, noting the definition of the upper and lower partitions is $M_i=\mathrm {sup}\{f(x) \lvert \text{P is a partition} \}$ and $m_i=\mathrm {inf}\{f(x)\lvert \text{P is a partition} \}$, this means $$M_i = f(x_{i-1}) = \frac{a}{1+\frac{(i-1)(e-1)}{n}} \quad \mathrm {and} \quad m_i = f(x_i) = \frac{a}{1+\frac{i(e-1)}{n}}$$

So the upper sum is $$U(f,P) = \sum_{i=1}^n M_i\Delta x = \sum_{i=1}^n \frac{e-1}{n+(i-1)(e-1)}$$ and the lower sum is $$L(f,P)=\sum_{i=1}^n m_i\Delta x=\sum_{i=1}^n\frac{e-1}{n+i(e-1)}$$

But I can't evaluate these sums. I'm not sure if my process is wrong, so I am getting difficult sums or what.

$\endgroup$
3
  • 1
    $\begingroup$ The partial sums are a digamma function. You also can factor $\sum_n\frac a{b n+c}=\frac ab\sum_n \frac1{n+\frac cb}$ $\endgroup$
    – Тyma Gaidash
    Commented Aug 5, 2022 at 17:00
  • 1
    $\begingroup$ Use the partition with points in geometric progression. Thus define $x_i=e^{i/n} $ for $i=0,1,2,\dots, n$ and then evaluate upper and lower sums. $\endgroup$
    – Paramanand Singh ♦
    Commented Aug 5, 2022 at 17:40
  • $\begingroup$ @Paramanand Singh Thank you! Solved it! $\endgroup$
    – Lennon Kirby
    Commented Aug 5, 2022 at 18:51

0

Reset to default

You must log in to answer this question.