Solve $y = \frac{x^6}{6} + \frac{1}{16x^4}$ by completing a perfect square, with a given domain.

Question

Problem:

$$y = \frac{x^6}{6} + \frac{1}{16x^4} \text{ for } 4 \le x \le 25$$

Solve by completing the square and use regular anti-derivatives.

Not even sure where to begin. I'm taking online classes and the professor is useless. The only completing the square I know of is when given a trinomial, which I do not have. I'm not looking for someone to do the work or give me the answer, but to just help me out with the steps so I can do it on my own.

Answer 1

The exercise is about the arc length of that curve.

By definition, you have that the arc length is given by

$$\int_a^b \sqrt{\left(\frac{\text{d}y}{\text{d}x}\right)^2 + 1}\ \text{d}x$$

In your case, the extrema $a$ and $b$ regimen:

$$a = 4 ~~~~~~~ b = 25$$

Your $y$ is given too, so you have to take its derivative:

$$\frac{\text{d}y}{\text{d}x} = x^5 - \frac{1}{4x^5}$$

Hence

$$\left(\frac{\text{d}y}{\text{d}x}\right)^2 = x^{10} + \frac{1}{16x^{10}} - \frac{1}{2}$$

So you have to evaluate this:

$$\int_4^25 \sqrt{x^{10} + \frac{1}{16x^{10}} - \frac{1}{2} + 1}\ \text{d}x = \int_4^25 \sqrt{x^{10} + \frac{1}{16x^{10}} + \frac{1}{2}}\ \text{d}x$$

Evaluation

This is actually easy, because what you have under the square root is nothing but a perfect square like the previous one, but with a + sign instead of a minus sign:

$$x^{10} + \frac{1}{16x^{10}} + \frac{1}{2} \equiv \left(x^5 + \frac{1}{4x^5}\right)^2$$

Hence the square and the square root are simplified, obtaining:

$$\int_4^{25} \left(x^5 + \frac{1}{4x^5}\right)\ \text{d}x$$

Which is trivial and you shall do it by yourself.

Complete Solution

In case you are completely lost, here it is.

You can split the integral in two parts, and the first integral is

$$\int_4^{25} x^5\ \text{d}x = \frac{x^6}{6}\bigg|_4^{25} = \frac{25^6}{6} - \frac{4^6}{6} = \frac{81378843}{2}$$

The second one:

$$\int_4^{25} \frac{1}{4x^5}\ \text{d}x = \frac{1}{4}\int_4^{25} \frac{1}{x^5}\ \text{d}x = -\frac{1}{16x^4}\bigg|_4^{25} = \frac{390369}{400000000}$$

So finally:

$$\frac{81378843}{2} + \frac{390369}{400000000} = \frac{16275768600390369}{400000000} \approx 4.06\cdot 10^7$$

Answer 2

Hint: From the comments, it sounds as though you're being asked to find the arc length of this curve between the given values. This is $$\int_4^{25}\sqrt{\left(\frac{dy}{dx}\right)^2+1}\,dx = \int_4^{25}\sqrt{\left(-\frac{1}{4x^5} + x^5\right)^2+1}\,dx.$$ Now expand, and complete the square as the hint says.