IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Xavier4

Evaluate: $\int_0^1 \left(1-x^{2006} \right)^{\frac{1}{2007}} dx + \int_0^1 \left(1-x^{2007} \right)^{\frac{1}{2006}} dx$

hsbhatt

I made a monumental error. Actually it should be $\int_0^1 (1-x^{2006})^{\frac {1}{2007}} - \int_0^1 (1-x^{2007})^{\frac {1}{2006}}$

I guess allamraju has the solution for this. So, I will wait one day for him to post it. Otherwise I will go ahead.

Meanwhile please accept 1000 apologies for the careless mistake.

Xavier4

Sir, please reply us.we all are waiting for ur reply

hsbhatt

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 \\ \\ \text{Let} \ x = \left(1-t^{2007}\right )^{\frac{1}{2006}} \\ \\ \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 \\ \\ \text{Now apply integration by parts. You get} \\ \\ I = -|t\left(1-t^{2007}\right )^{\frac{1}{2006}}|_0^1 + \int_0^1 \left(1-t^{2007}\right )^{\frac{1}{2006}} dt \\ \\ = \int_0^1 \left(1-t^{2007}\right )^{\frac{1}{2006}} dt \\ \\ =\int_0^1 \left(1-x^{2007}\right )^{\frac{1}{2006}} dx \\ \\ \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 \\ \\ \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 \\ \\ \text{Finis !!} \\ \\ \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 $

anchitsaini

great sir , dunno why people are not appreciating your efforts nowadays !!! :( :(

pranay_robot

gr8888 sir ... thanks a lot for posting the solution

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