$\int_{1}^{2}\frac{e^x+e^{4/x}}{x}dx$

Question

$I=\int_{1}^{2}\frac{e^x+e^{4/x}}{x}dx$

For $ x \in (1,2), \frac{e^x+e^{4/x}}{x}>e^x$

$\Rightarrow I=\int_{1}^{2}\frac{e^x+e^{4/x}}{x}dx$ >$\int_{1}^{2}e^xdx$

$\Rightarrow I>e^2-e$

And $I<e+e^4$ by considering a rectangle

Is there a better way we can approximate the integral more efficiently by assuming some function like $e^x$ which can be understood by a highschool student?

Answer 1

Is there a better way we can approximate the integral more efficiently by assuming some function like $e^x$ which can be understood by a highschool student?

The exact result is expressed in terms of special functions beyond such students' education. As the comments have suggested, we can numerically approximate by whatever methods the curriculum covers. To improve the approximation, divide the integration range into more strips (see e.g. here). But it's a job for a computer.

Answer 2

If you want approximations, think about Taylor series centered at the mid point. This would give for the integrand $$\frac{2}{3} \left(e^{3/2}+e^{8/3}\right)+\frac{2e^{3/2}}{27} \left(3-22 e^{7/6}\right) \left(x-\frac{3}{2}\right)+\frac{ e^{3/2}}{243} \left(45+712 e^{7/6}\right) \left(x-\frac{3}{2}\right)^2-$$ $$\frac{e^{3/2} \left(81+29584 e^{7/6}\right) }{6561}\left(x-\frac{3}{2}\right)^3+\left(\frac{35 e^{3/2}}{972}+\frac{371296 e^{8/3}}{59049}\right) \left(x-\frac{3}{2}\right)^4+O\left(\left(x-\frac{3}{2}\right)^5\right)$$ Now, integrate termwise using the bounds and get $$\int_{1}^{2}\frac{e^x+e^{4/x}}{x}dx \sim \frac{10615 e^{3/2}}{15552}+\frac{292126 e^{8/3}}{295245} \approx 17.2989$$ while the exact result would be $ \approx 17.7358$

Answer 3

As @J.G. said we'll need in our answer some special functions so we can get the exact result: $$\int_1^2 \frac{e^x+e^{4/x}}x \mathrm{d}x=\int_1^2 \frac{e^x}x \mathrm{d}x +\int_1^2 \frac{e^{4/x}}x \mathrm{d}x \ \ \ \ \ \text{linearity}$$ We can observe that : $$\mathrm{Ei}(x)=\int\frac{e^x}x\mathrm{d}x$$ For the other one we can with a simple substitution get the following:$ \ u=\frac{4}x \Rightarrow \mathrm{d}x=-\frac{x^2}4\mathrm{d}u $ $$-\int \frac{e^u}u \mathrm{d}u=-\mathrm{Ei}(u)=-\mathrm{Ei}\left(\frac{4}x\right)$$ And here's our antiderivative: $\displaystyle \mathrm{Ei}(x)--\mathrm{Ei}\left(\frac{4}x\right)$ \begin{align}\int_1^2 \frac{e^x+e^{4/x}}x \mathrm{d}x&=\mathrm{Ei}(x)-\mathrm{Ei}\left(\frac{4}x\right)\bigg\vert_1^2\\ &=\mathrm{Ei}(2)-\mathrm{Ei}\left(\frac{4}2\right)-\mathrm{Ei}(1)+\mathrm{Ei}(4)\\ &=\mathrm{Ei}(4)-\mathrm{Ei}(1)\\ &\approx 17.735756 \end{align} And that's what @ClaudeLeibovici considered as the exact result!