How to evaulate this integral using Newtons binomial theorem?

Question

Evaluate the following definite integral: $$\int_0^{0.5}(3-5x)^{\frac{4}{3}} \ \mathrm dx$$

I'm working with this integral and I want to solve it using Newton's Binomial Theorem. Not sure how to go about that. Could use some tips/help

Answer 1

Normal method

Let $u=3-5x$. Then, $\mathrm du = -5 \ \mathrm dx$.

Also, $x=\dfrac{3-u}5$. Then, $\mathrm dx = -\dfrac15\ \mathrm du$.

When $x=0$, $u=3$; when $x=0.5$, $u=0.5$.

Then:

$$\begin{array}{rcl} \displaystyle \int_0^{0.5}(3-5x)^{\frac{4}{3}} \ \mathrm dx &=& \displaystyle \int_3^{0.5}u^{\frac{4}{3}} \left(-\dfrac15 \ \mathrm du\right) \\ &=& \displaystyle \dfrac15 \int_{0.5}^3 u^{\frac{4}{3}} \ \mathrm du \\ &=& \displaystyle \dfrac15 \left(\dfrac{3u^{\frac73}}{7}\right)_{0.5}^3\\ &=& \displaystyle \dfrac3{35} \left(9\sqrt[3]3-0.25\sqrt[3]{0.5}\right)\\ &=& \displaystyle \dfrac3{280} \left(72\sqrt[3]3-\sqrt[3]{4}\right)\\ \end{array}$$

---

Binomial theorem

$$\begin{array}{rcl} \displaystyle \int_0^{0.5}(3-5x)^{\frac{4}{3}} \ \mathrm dx &=& \displaystyle 3\sqrt[3]3 \int_0^{0.5} \left(1-\dfrac53x\right)^{\frac{4}{3}} \ \mathrm dx \\ &=& \displaystyle 3\sqrt[3]3 \int_0^{0.5} \left(\sum_{n=0}^\infty \dbinom{4/3}{n} \left(-\dfrac53x\right)^n\right) \ \mathrm dx \\ &=& \displaystyle 3\sqrt[3]3 \left(\sum_{n=0}^\infty \int_0^{0.5} \dbinom{4/3}{n} \left(-\dfrac53x\right)^n \ \mathrm dx\right) \\ &=& \displaystyle -\dfrac{9\sqrt[3]3}5 \left(\sum_{n=0}^\infty \dfrac1{n+1} \dbinom{4/3}{n} \left(-\dfrac56\right)^{n+1}\right) \\ &=& \displaystyle -\dfrac{9\sqrt[3]3}5 \left(\sum_{n=0}^\infty \dfrac37 \dbinom{7/3}{n+1} \left(-\dfrac56\right)^{n+1}\right) \\ &=& \displaystyle -\dfrac{27\sqrt[3]3}{35} \left(\sum_{n=1}^\infty \dbinom{7/3}n \left(-\dfrac56\right)^n\right) \\ &=& -\dfrac{27\sqrt[3]3}{35} \left[\left(1-\dfrac56\right)^{\frac73} - 1\right]\\ &=& -\dfrac{27\sqrt[3]3}{35} \left[\dfrac1{6^{\frac73}} - 1\right]\\ &=& \dfrac{3}{140\sqrt[3]2} \left[36\sqrt[3]6 - 1\right]\\ &=& \dfrac{3}{280} \left[72\sqrt[3]2 - \sqrt[3]4\right]\\ \end{array}$$

Answer 2

To use the binomial theorem for fractional exponents, you should make one of the terms inside the parentheses be $1$, in which case the other has to be less than 1 in magnitude in order to guarantee that the series converges. In this case notice that $|5x| \leq 2.5<3$ on the domain of interest, so we write

$$(3-5x)^{4/3}=3^{4/3}(1-(5/3)x)^{4/3}.$$

Once you are in that form you just plug in the formula $(1+x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n} x^n$ where ${\alpha \choose n}$ is defined to be $\frac{\alpha (\alpha-1) \cdots (\alpha-n+1)}{n!}$. You can then integrate term-by-term to get a series representation of the integral. Finally you can rearrange this series back into binomial theorem form in order to get a closed form solution.

This is much more difficult than just taking a substitution and using the power rule, but it does work in the end.