I need to get the length of a curve which equation is :
$$y= (4-x^\frac{2}{3})^\frac{3}{2}$$
I need to find the length using the method :
$$L=\int_a^b \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}$$
So I started by evaluating dy/dx which gave me :
$$-\frac{\sqrt{4-x^\frac{2}{3}}}{\sqrt[3]{x}}$$
Then I integrated using the method stated before which gave me at end :
$$x\sqrt{2}\left(1-\frac{2}{x^\frac{2}{3}}\right)^\frac{3}{2}$$
Last step, I tried to evaluate the integral from 0 to 8 which is the domain of the curve... But I realised my integral was unsolvable for 0 So I guess I made a mistake somewhere...
What Am I missing here ?
Solution
My mistake was that I kept the minus sign even when elevating to 2.
The integral was actually easy to resolve without that minus sign...
$$L=\int_a^b \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}dx$$
$$= \int_a^b \sqrt{ 1 +\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}} + 1}dx$$
$$= \int_a^b \frac{2}{\sqrt[3]{x}}dx$$
$$= 2 \int_a^b \frac{1}{\sqrt[3]{x}}dx$$
$$= 3x^\frac{2}{3} + C$$
Now I need to evaluate from 0 to 8... Which equals to 12u.
Let us simplify $1+\left(\frac{dy}{dx}\right)^2$. From your correct expression for the derivative, we get $$1+\left(\frac{dy}{dx}\right)^2=1+\frac{4-x^{2/3}}{x^{2/3}}.$$ Bring the expression on the right to the common denominator $x^{2/3}$. There is nice cancellation, and the problem collapses.
Remarks: $1.$ In principle there is an issue at $0$, since our integrand blows up there. But it is a convergent improper integral.
$2.$ There was some incorrect algebra in your calculation.