Question

any ideas on how to evaluate
\[\int_{0}^{1} \sqrt[7]{1-x^3}-\sqrt[3]{1-x^7} dx\]

Answer 1

\[\int_{0}^{1} (\sqrt[7]{1-x^3}-\sqrt[3]{1-x^7} )dx\]

Answer 2

its in the james stewart 6th edition calculus book the poblem plus section at the end of chapter 8 if anyone is interested

Answer 3

and actually i wrote it backwards oh well

Answer 4

stewart? never heard of it

Answer 5

lol
sorry for the delayed response my computer died

Answer 6

in fact i have no idea how to do it. and i have only the stupid "essential" version with me

Answer 7

we need the power of zarkon!

Answer 8

Cogs turning ... I notice first that the two functions in the integrand are inverse of each other on [0,1]. But there's no magical theorem about the integral of (f -inv.f). None the less, context would be good--what is the author discussing immediately before this exercise?

I also note that the function has one zero on (0,1); call it a. I've been trying to prove the integral from 0->a equals the negative of the integral a->1 without success yet.

Answer 9

well its the problem plus section so anywhere from integration by parts, parital fractions, trig substitution, substitution
there is also one thing the author used that i thought would be useful but had no luck
in an example he proved
\[\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x) dx\]

Answer 10

ok -- tv watching + problem solving, usually equals zero, but I'll give it a go.

Answer 11

ok there is no hurry james
i was thinking about giving it as a bonus assignment to my students
but if i can't figure it out then i shouldn't make them try to figure it out
lol
i was just stumped by it
i will also keep thinking on it

Answer 12

\[\int_{0}^{1} (1-x^3)^{1/7}-{(1-x^7)^{1/3}} dx\]
the xs get ya

Answer 13

i wish i could remember, but there is something to do with a substitution with exponents

Answer 14

you find a common denominator or some such and it goes z^(?) = some such thing

Answer 15

\[z^7=1-x^3\]
\[(1-z^7)^{1/3}=x\]
\[7z^6 dz=-3x^2 du\]

Answer 16

\[7z^6 dz=-3(1-z^7)^{2/3} du\]
\[\frac{7z^6}{-3(1-z^7)^{2/3}} dz= dx\]

Answer 17

...wish i could recall if this was the method im thinking of or if im just inventing it lol

Answer 18

i don;t see anything wrong with it but i'm not sure if we can use it

Answer 19

\[\int\frac{7z^6}{-3z(1-z^7)^{2/3}} dz\]
\[\int\frac{-7z^6}{3z(1-z^7)^{2/3}} dz\]
im not sure either :)

Answer 20

still looks nasty

Answer 21

this shouldn't be hard
lol
its just cal 2
cal 2>me :(

Answer 22

7*3 = 21 so it was prolly something akin to z^{21} youd use or some such

Answer 23

speak lagrange :)

Answer 24

Ok. First,

LEMMA
Suppose f an increasing continuous function on [a,b]. Then
\[\int\limits_{a}^{b} f^{-1} \ = \ b f^{-1}(b) - af^{-1}(a) - \int\limits_{f^{-1}(a)}^{f^{-1}(b)} f \ .\]

PROOF
Let
\[P = \{ t_0, t_1, ..., t_n \}\]
be a partition of [a,b] and let
\[P' = \{ f^{-1}(t_0), ..., f^{-1}(t_n) \} .\]

Then carefully writing out the sums with a diagram, we see that if L is a lower Riemann sum and U the upper sum,

\[L(f^{-1}, P) + U(f, P') = \ b f^{-1}(b) - af^{-1}(a)\]

The result now follows. Note also that we can obtain the same result with f decreasing on [a,b] by swapping L and U above. qed.

Now, given all that, let
\[f(x) = \sqrt[7]{1-x^3} \ \hbox{ and hence } \ f^{-1}(x) = \sqrt[3]{1-x^7}\]

Consider
\[I = \int\limits_{0}^{1} f^{-1} = 1.f^{-1}(1) - 0.f^{-1}(0) - \int\limits_{f^{-1}(0)}^{f^{-1}(1)} f\]
\[ = 0 - 0 - \int\limits_{1}^{0} f = \int\limits_{0}^{1} f \]
and therefore for our f
\[\int\limits_{0}^{1} (f^{-1} - f\ ) = 0 \ .\]
Hence
\[\int\limits_{0}^{1} (\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3} \ ) = 0 \ .\]

Answer 25

wow james

Answer 26

amazing!!!

Answer 27

james is the man

Answer 28

totally cute!
too advance for cal 2 students though lol

Answer 29

@Myininaya: where do you teach?

Answer 30

its a small college in little rock Arkansas

Answer 31

The result can also be shown by just using integration by parts

for

\[\int\limits_{a}^{b}f^{-1}(x)dx\]

let \[u=f^{-1}(x)\text{ and }dv=dx\]

and result follows fairly quickly.

Answer 32

Really? I don't think it does, as du is messy.

Answer 33

\[du=\frac{1}{f'(f^{-1}(x))}dx\text{ and }v=x\]

\[\int\limits_{a}^{b}f^{-1}(x)dx=\left.xf^{-1}(x)\right|_{a}^{b}-\int\limits_{a}^{b}\frac{x}{f'(f^{-1}(x))}dx\]

\[=bf^{-1}(b)-af^{-1}(a)-\int\limits_{a}^{b}\frac{x}{f'(f^{-1}(x))}dx\]

\[u=f^{-1}(x) \Rightarrow f(u)=x\]


since \[du=\frac{1}{f'(f^{-1}(x))}dx\]

we have \[=bf^{-1}(b)-af^{-1}(a)-\int\limits_{u(a)}^{u(b)}f(u)du\]

\[=bf^{-1}(b)-af^{-1}(a)-\int\limits_{f^{-1}(a)}^{f^{-1}(b)}f(u)du\]

call x=u

\[=bf^{-1}(b)-af^{-1}(a)-\int\limits_{f^{-1}(a)}^{f^{-1}(b)}f(x)dx\]

Answer 34

Got it.

Answer 35

james+zarkon=world domination by math