Hello people, I was just hitting my chapter review exercises, but one of them I thinked of if for hours and I cannot simplify it in a manner that makes it resolvable by hand.
#Unsolvable Integral in the book?
fair estuary
minor ospreyBOT
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The task is to find the arc length in interval [1; 2] for the following function
$y=\frac{x^5}{10}+\frac{x^{-3}}{6}$
sharp ginkgoBOT
Fractal Verse
fair estuary
Now, I'll tell you my steps to show you where I got and where I believe it cannot be further simplified
I applied the standard formula for arc length over an interval, which is the following
$L_{[a;b]}=\int_{a}^{b}\sqrt{1+\left[ \frac{dy}{dx} \right]^2}dx$
sharp ginkgoBOT
Fractal Verse
fair estuary
And my calculations steps are the following
$L=\int_{1}^{2}\sqrt{1+\left[ \frac{d}{dx}\left( \frac{x^5}{10}+\frac{x^{-3}}{6} \right)\right]^2}dx=\int_{1}^{2}\sqrt{1+\left[ \frac{x^4}{2}+\frac{x^{-4}}{2}\right]^2}dx=\int_{1}^{2}\sqrt{1+\frac{1}{4}\left[ x^4+x^{-4}\right]^2}dx=\int_{1}^{2}\sqrt{1+\frac{1}{4}(x^8-2-x^{-8})}dx=\int_{1}^{2}\sqrt{\frac{x^8}{4}+\frac{x^{-8}}{4}+\frac{1}{2}}dx$
sharp ginkgoBOT
Fractal Verse
fair estuary
I don't believe this stuff can be solved without a calculator? Am I doing something wrong, because this exercise wasn't put in the CAS solving section.
the calculator can solve it, but I have no clue
lean widget
My sympy is simplifying it to something much simpler
The sqrt is gone in my sympy simplification
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Oh wait, I can simplify out the entire square root
devout dirge
yeah
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$\int_1^2 \frac{x^8+1}{2x^4}dx$
sharp ginkgoBOT
Fractal Verse
fair estuary
Thank you so much guys. I can now split it in two integrals and solve easily