Alright I integrate x^4+2/x^4 to get -2/3x^3 + x^5/5 now how do evaluate it at 0 to 1 and prove if it diverges?
www.gummy-stuff.org/Calculus/calculus-3.pdf- for better explanation
if you integrate it \[x^{5}/5-2/3x^{3}\]
Is it a definite integral?
[(-2/3x^3 + x^5/5) - (-2/3x^3 + x^5/5)] Where in the left hand side you plug in 1 and in the right hand side you plug in 0
\[\int\limits_{0}^{1}x^4+2/x^4 dx\]
That's basically what they are asking you to do.
@Romero that is what they are asking me to solve it's basically if it converges or diverges
it's an improper integral:\[\left[\begin{matrix}\frac{x^5}{5} - \frac{2}{3x^3} \end{matrix}\right]^{1}_{a}\] \[=\frac{1}{5} - \frac{2}{3} - \frac{a^5}{5} + \frac{2}{3a^3}\] now as a tends to zero we can say that the last term diverges to infinity thats how i learnt it anyway :)
yup - thats correct
Well when you evaluate the ingetral from 0 to 1 and you get an exact value then it converges but if in this case since you have x in the denomenator then it diverges.
fun fact with improper integrals: if you find the volume enclosed from 1 to +infinity when the function y = 1/x is rotated about the x axis you'll get it converging to pi but if you find the surface area of this function from 1 to +infinity we find it diverges. so if it a bucket of paint could be built in this shape we wouldn't be able to paint the outside with the paint we could fit inside
Torricelli's trumpet =)
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