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• Of course the easiest way is with technology: Maple tells me that ... 9.000042868 which works for me. A more difficult option would be to get enough terms in
Message 1 of 4 , Jan 1, 2003
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Of course the easiest way is with technology: Maple tells me that

> evalf(int((x^4+1)^(1/4),x=0..3)+int((x^4-1)^(1/4),x=1..3));

9.000042868

which works for me.

A more difficult option would be to get enough terms in the Taylor
series so that the error was gauranteed to be sufficiently small, and
then evaluate each of these. Since the interval is so large, you
would need a large number of terms, so that sounds way too laborious.

I am happy with the technology answer.........

--- In mathforfun@yahoogroups.com, j031p <no_reply@y...> wrote:
> I have received this problem about a week ago and I don't know how
> to really attack this problem. I wonder if you guys can give me a
> hand. Thanks
>
> 3 3
> 9 < $(x^4+1)^1/4 dx +$ (x^4-1)^1/4 dx < 9.0001
> 0 1
>
> $= integration notation > > Joel > > PS I tried miscellaneous substitution but that hasn't really worked > for me either. • ... Let (*) be the center expression in your problem. Let f(x) = (x^4 - 1)^{1/4}. Since f is strictly increasing with f(1) = 0, we have that int_1^3 {f(x) dx} Message 1 of 4 , Jan 1, 2003 View Source --- In mathforfun@yahoogroups.com, j031p <no_reply@y...> wrote: > I have received this problem about a week ago and I don't know how > to really attack this problem. I wonder if you guys can give me a > hand. Thanks > > 3 3 > 9 <$ (x^4+1)^1/4 dx + $(x^4-1)^1/4 dx < 9.0001 > 0 1 > >$ = integration notation
>
> Joel
>
> PS I tried miscellaneous substitution but that hasn't really worked
> for me either.

Let (*) be the center expression in your problem.

Let f(x) = (x^4 - 1)^{1/4}. Since f is strictly increasing with f(1)
= 0, we have that

\int_1^3 {f(x) dx} = 3*f(3) - \int_0^{f(3)} {f^{-1}(y) dy}.

But f^{-1}(y) = (y^4 + 1)^{1/4}, so

(*) = 3*f(3) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
= 9 - 9 + 3*f(3) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
= 9 - 3*(3 - f(3)) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
= 9 + \int_{f(3)}^3 {((x^4 + 1)^{1/4} - 3) dx}.

Finally, (x^4 + 1)^{1/4} is concave up, so the integral above is
bounded above by the area of the triangle with base 3 - f(3) and
height (3^4 + 1)^{1/4} - 3, i.e.

9 < (*) < 9 + (3 - 80^{1/4})*(82^{1/4} - 3)/2.

Now, get out the calculator to see

9 < (*) < 9.00004287
• ... how ... worked ... (1) ... Plugging in a few more digits, (*)
Message 1 of 4 , Jan 1, 2003
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--- In mathforfun@yahoogroups.com, jason1990 <no_reply@y...> wrote:
> --- In mathforfun@yahoogroups.com, j031p <no_reply@y...> wrote:
> > I have received this problem about a week ago and I don't know
how
> > to really attack this problem. I wonder if you guys can give me a
> > hand. Thanks
> >
> > 3 3
> > 9 < $(x^4+1)^1/4 dx +$ (x^4-1)^1/4 dx < 9.0001
> > 0 1
> >
> > \$ = integration notation
> >
> > Joel
> >
> > PS I tried miscellaneous substitution but that hasn't really
worked
> > for me either.
>
> Let (*) be the center expression in your problem.
>
> Let f(x) = (x^4 - 1)^{1/4}. Since f is strictly increasing with f
(1)
> = 0, we have that
>
> \int_1^3 {f(x) dx} = 3*f(3) - \int_0^{f(3)} {f^{-1}(y) dy}.
>
> But f^{-1}(y) = (y^4 + 1)^{1/4}, so
>
> (*) = 3*f(3) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
> = 9 - 9 + 3*f(3) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
> = 9 - 3*(3 - f(3)) + \int_{f(3)}^3 {(x^4 + 1)^{1/4} dx}
> = 9 + \int_{f(3)}^3 {((x^4 + 1)^{1/4} - 3) dx}.
>
> Finally, (x^4 + 1)^{1/4} is concave up, so the integral above is
> bounded above by the area of the triangle with base 3 - f(3) and
> height (3^4 + 1)^{1/4} - 3, i.e.
>
> 9 < (*) < 9 + (3 - 80^{1/4})*(82^{1/4} - 3)/2.
>
> Now, get out the calculator to see
>
> 9 < (*) < 9.00004287

Plugging in a few more digits,

(*) < 9.00004286888

Using convexity, we can also get a lower bound. The function

(x^4 + 1)^{1/4} - 3

is bounded below by its tangent line at x = 3, which is given by

y = (x - f(3))/f'(3).

Integrating this from 0 to f(3) and plugging it into the calculator
gives

9.00004286642 < (*) < 9.00004286888
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