Integral Involving Irrational values

[courtrigrad]

Hello all

If we want to calculate the definite integral $$\int^b_a x^{\alpha}$$ for any irrational value of $$\alpha$$ where $$0<a<b$$ do we use the Mean Value Theorem? Would $$\alpha$$ be represented as a limit of a sequence of rational numbers $$\alpha = \lim_{x\rightarrow \infty} \alpha_n$$ and $$\alpha$$ is not equal to -1. Hence $$x^{\alpha} = \lim_{x\rightarrow \infty} x^{\alpha_n}$$ So we can always find a number such that $$|x^{\alpha} - x^{\alpha_n}| < \epsilon$$ (how do we prove this)?.

Now $$f(x) = x^\alpha$$ and $$g(x) = x^{\alpha_n}$$. Now applying the Mean Value Theorem for Integral Calculus we get:

$$-\epsilon(b-a) + \int^b_a x^{\alpha_n} \ dx < \int^b_a x^{\alpha} \ dx < \int^b_a x^{\alpha_n} + \epsilon(b-a)$$

We know that $$\int^b_a x^{\alpha} \ dx = \frac {1}{\alpha +1}(b^{\alpha +1} - a^{\alpha+1})$$.

$$-\epsilon(b-a) + \frac{1}{\alpha_n +1}(b^{\alpha_n+1} - a^{\alpha_n+1}) < \int^b_a x^\alpha \ dx < \frac{1}{\alpha_n +1}(b^{\alpha_n+1} - a^{\alpha_n+1}) + \epsilon(b-a)$$. From here how do we receive

$$\int^b_a \ dx = \frac{1}{\alpha +1}(b^{\alpha+1} - a^{\alpha +1})$$?

Thanks

 

[vincentchan]

use
$$x^\alpha = e^{\alpha lnx}$$

 

[Hurkyl]

We could use the FTAC, because we know the derivative of x^&alpha;+1.

 

[HallsofIvy]

If y= x^α then ln(y)= α ln(x) so $\frac{1}{y}\frac{dy}{dx}= \alpha \frac{1}{x}$. That is, $\frac{dy}{dx}= \alpha\frac{1}{x}y= \alpha\frac{1}{x}x^{\alpha}= \alpha x^{\alpha-1}$

From that, it follows that $\frac{1}{\alpha+1}x^{\alpha+1}$ is an anti-derivative of $x^{\alpha;}$.

$\int^b_a x^{\alpha}= \frac{1}{\alpha+1}x^{\alpha+1}+ C$.