My, what large powers you have!

Calculus Level 3

The integral of $(x+1)(\frac{x}{2} - 1)^{10}$ with respect to $x$ can be expressed as $\frac{\alpha}{\beta} (\frac{x}{2}-1)^\gamma + \frac{\delta}{\epsilon} (\frac{x}{2}-1)^\omega + c$ where $c$ is the arbitrary constant. $gcd (\alpha, \beta) = gcd (\delta, \epsilon) =1$ and $\gamma > \omega$ .

All I need you to add up is $\delta$ and $\epsilon$ . What is the answer???

The answer is 17.

1 solution

Noel Lo
Apr 22, 2015

Let $u= \frac{x}{2} + 1$ so that $x=2u+2$ . Now we have $(2u+2+1)u^{10} \times \frac{dx}{du} = (2u+3)u^{10} \times 2 = 4u^{11} + 6u^{10}$ or $4(\frac{x}{2} - 1)^{11} + 6(\frac{x}{2} - 1)^{10}$ which upon integration gives us $\frac{4}{12} (\frac{x}{2} - 1)^{12} + \frac{6}{11} (\frac{x}{2} - 1)^{11}+c$ = $\frac{1}{3} (\frac{x}{2} - 1)^{12} + \frac{6}{11} (\frac{x}{2} - 1)^{11}+c$ .

Since $\gamma > \omega$ , $\gamma = 12, \omega = 11, \delta = 6, \epsilon = 11$ . So $\delta + \epsilon = 6+11 =\boxed{17}$ .

My, what large powers you have!

The answer is 17.

1 solution

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