Value of k in definite Integration

[juantheron]

Let \(\displaystyle p(x)=2x^6+4x^5+3x^4+5x^3+3x^2+4x+2.\) Let \(\displaystyle \displaystyle I_{k}=\int^{\infty}_{0}\frac{x^k}{p(x)}dx\)

where \(\displaystyle 0<k<5.\) Then value of \(\displaystyle k\) for which \(\displaystyle \displaystyle I_{k}\) is smallest.

 

[jvicens]

I would first clarify what the forum poster means by "smallest" - do they mean the smallest numerical value, or the smallest area under the curve? Assuming they mean the smallest numerical value, I would approach the problem by first finding the indefinite integral of p(x), which is:

\displaystyle \int p(x)dx = \frac{1}{7}x^7 + \frac{2}{3}x^6 + \frac{3}{2}x^5 + \frac{5}{4}x^4 + x^3 + 2x^2 + 2x + C

Then, I would use the given formula for I_k to find the definite integral from 0 to infinity:

\displaystyle I_k = \left[\frac{1}{7}x^{k+7} + \frac{2}{3}x^{k+6} + \frac{3}{2}x^{k+5} + \frac{5}{4}x^{k+4} + x^{k+3} + 2x^{k+2} + 2x^{k+1}\right]_{0}^{\infty}

Since the lower limit is 0, all terms with an exponent greater than or equal to 1 will evaluate to 0. This leaves us with:

\displaystyle I_k = \left[\frac{1}{7}x^{k+7} + \frac{2}{3}x^{k+6} + \frac{3}{2}x^{k+5} + \frac{5}{4}x^{k+4}\right]_{0}^{\infty}

Since k is a positive value, as x approaches infinity, the terms with the highest exponents will dominate and give us an infinite value. Therefore, in order for I_k to be the smallest numerical value, we want the highest exponent to be as small as possible. This would occur when k+4 = 0, or when k = -4. However, the given condition is that 0<k<5, so the smallest value for k would be 0.

In conclusion, the value of k for which I_k is smallest is 0.