Approximating $\int_2^4 e^{\frac{1}{x}} dx$

Question

Estimate the integral precisely up to the error of $0.001$:$$\int_2^4 e^{\frac{1}{x}} dx$$

I started by dividing the segment $[2,4]$ into some pieces $(n=5)$ as in the following figure:

$\qquad\qquad\quad$

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Now, I have $$\Delta x = \frac{4-2}{5} = 0.4$$ and therefore, $$S \approx \sum_{i=1}^5 f(x_i)(x_i - x_{i-1}) = \Delta x \sum_{i=1}^5 e^{\frac{1}{x_i}}$$ where $x_i = 2 + 0.4i$.

Then, approximating the area boils down to approximating the exponential function at $5$ points. What is the best way to proceed? And how do I know that the final result will give the error at most $0.001$?

Answer 1

The Midpoint Rule error is given by

$$\tag 1 |E_n| \le \left|\dfrac{(b-a)^3}{24 n^2}\right | |\max_{2 \le x \le 4}f''(x^*)|$$

The second derivative of $f(x) = e^{1/x}$ is

$$f''(x)= \dfrac{e^{1/x} (2 x + 1)}{x^4}$$

So, we want to find the max of $|f''(x)|$ over the interval $[2,4]$

$$\displaystyle \max_{2 \le x \le 4}\left| \dfrac{e^{1/x} (2 x + 1)}{x^4}\right| = \dfrac{5 \sqrt{e}}{16} ~ \mbox{at}~ x = 2$$

Substituting everything we have into $(1)$

$$|E_n| \le \dfrac{(4-2)^3}{24 n^2} \left(\dfrac{5 \sqrt{e}}{16}\right) \le 0.001$$

Can you continue by solving that for $n$? You should find

$$n \ge 13.105$$

We would choose an integer at $n = 14$.

Note that that is an upper bound and can be way off.

In practice, using trial and error gives eight intervals for an absolute error $< 0.001$

$$2.83456$$

Using WA, the closed form is ugly, and the numerical result is

$$\displaystyle \int_2^4 e^{1/x} dx \approx 2.835422294875796952$$

Update If you want to see the intermediate Midpoint Rule steps above, use this calculator.

Answer 2

$$\exp(1/x)=\sum_{n=0}^\infty\frac{x^{-n}}{n!}$$ $$\int\exp(1/x)dx=\sum_{n=0}^\infty\frac{x^{1-n}}{(1-n)n!}+C$$ then sub in your limits and take a few terms, so thats one option

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$$\int_2^4e^{1/x}dx=\int_{1/4}^{1/2}\frac{e^u}{u^2}du=\int_{1/4}^{1/2}\sum_{n=0}^\infty\frac{u^{n-2}}{n!}du$$ so you have two choices here