Question

having some trouble with this one

Answer 1

i turned it into exponential form and then i'm not really sure where to go after that

Answer 2

what have you got so far @acidhellscream

Answer 3

1+s^(1/4)

Answer 4

\[\int\limits_1^{\sqrt[4]2}\frac{s^2+\sqrt[4]s}{s^2}\cdot\text ds\]
\[=\int\limits_1^{\sqrt[4]2}\frac{s^2}{s^2}+\frac{s^{1/4}}{s^2}\cdot\text ds\]
\[=\int\limits_1^{\sqrt[4]2}s^{2-2}+s^{1/4-2}\cdot\text ds\]\[=\int\limits_1^{\sqrt[4]2}1+s^{-7/4}\cdot\text ds\]

Answer 5

ahh i forgot to split the it into the sums. brb im going to see if i can make more progress

Answer 6

\[\sqrt[4]{2}-4/3\sqrt[16]{2}-1/3\]

is this the final answer? or am i at least close?

Answer 7

thats a 16th root by the way its really small

Answer 8

First term is correct..

Check for second and last term..

Answer 9

Last term should be positive I think..

Answer 10

\[\large \left| s - \frac{4}{3}s^{\frac{-3}{4}} \right|_{1}^{\sqrt[4]{2}} \implies \sqrt[4]{2}- \frac{4}{3} \cdot 2^{\frac{1}{4} \cdot \frac{-3}{4}} -1 + \frac{4}{3}(1)^{\frac{-3}{4}}\]

\[\large \sqrt[4]{2} -\frac{4}{3} \sqrt[16]{2^{-3}} + \frac{1}{3}\]

Answer 11

haha oops i actually had that -3 in the radical i just wrote it small and didnt copy it over! thank you so much for your help!

Answer 12

Welcome..

Be careful for signs next time..