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hsbhatt (5772)

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Olaaa!! Perrrfect answer. 1086  [1257 rates]

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There are three ways to go about this, presented in increasing order of rigour:


1. Graphical solution.


First look at the two integrands: f(x) = \left(1-x^{2006}\right )^{\frac{1}{2007}} \ \text{and} \ g(x) = \left(1-x^{2007}\right )^{\frac{1}{2006}}


The main thing is to recognise one to be the inverse of the other.


Now, draw a tentative graph of, say, f(x). It will resemble the sine curve.


To get at the graph of g(x) is simple, you have to make the y-axis the x-axis. Since f(0) = g(0) =1 and f(1) = g(1) = 0, you can easily see that the area under the graph is the same for both.  \text{Hence} \ \int_0^1 \left(1-x^{2006}\right )^{\frac{1}{2007}} dx - \int_0^1 \left(1-x^{2007}\right )^{\frac{1}{2006}} dx = 0


2. Logical solution:


This depends on two facts:


1. f(x) and g(x) are inverses of each other


2. The domain and range of f(x) (and hence of g(x)) is [0,1]. For every t1 belonging to [0,1] we can find t2 belonging to [0,1] such that f(t1) = g(t2).


From this you can easily deduce that the two integrals have the same value


3. No-nonsense solution:


\text{Consider} \ I = \int_0^1 \left(1-x^{2006}\right )^{\frac{1}{2007}} dx \\ \\<br/>\text{Let} \ x = \left(1-t^{2007}\right )^{\frac{1}{2006}} \\ \\<br/>\text{Then} \ I = -\int_0^1 t \ d  \left(1-t^{2007}\right )^{\frac{1}{2006}}  =  -\int_0^1 t \ \frac{d  \left(1-t^{2007}\right )^{\frac{1}{2006}}}{dt} \ dt \\ \\<br/>\text{Now apply integration by parts. You get} \\ \\<br/>I = -|t\left(1-t^{2007}\right )^{\frac{1}{2006}}|_0^1 + \int_0^1 \left(1-t^{2007}\right )^{\frac{1}{2006}} dt \\ \\<br/>= \int_0^1 \left(1-t^{2007}\right )^{\frac{1}{2006}} dt \\ \\<br/>=\int_0^1 \left(1-x^{2007}\right )^{\frac{1}{2006}} dx \\ \\<br/>\text{Thus} \ \int_0^1 \left(1-x^{2006}\right )^{\frac{1}{2007}} dx  = \int_0^1 \left(1-x^{2007}\right )^{\frac{1}{2006}} dx  \\ \\<br/>\text{or} \ \int_0^1 \left(1-x^{2006}\right )^{\frac{1}{2007}} dx  - \int_0^1 \left(1-x^{2007}\right )^{\frac{1}{2006}} dx = 0 \\ \\<br/>\text{Finis !!} \\ \\<br/>\text{In general you can remember that} \ \int_0^1 \left(1-x^{a}\right )^{\frac{1}{b}} dx  - \int_0^1 \left(1-x^{b}\right )^{\frac{1}{a}} dx = 0<br/>


 


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