Arc length of definite integral Let $F$ be defined for $x\geq$ $1$ where
$\displaystyle F(x) = \int_{1}^{x} \sqrt{25t^3-1}$ $dt$
Let $K$ be the arclength $y = F(x)$ for $1 \leq x \leq 4$. Find the arclength.
Now I'm attempting to make use of the formula $\displaystyle K = \int_{1}^{4} \sqrt{1+(F'(x))^2}$, but I'm struggling to find the derivative of the definite integral. I thought deriving the integral would simply yield
$F(x)-F(1)$ since the derivative cancels the integral. 
By doing this I end up with $\sqrt{25x^3-1} - \sqrt{24}$
But I can already tell that something seems incorrect here, as the next step involves raising this expression to the power of 2 and then putting it into the formula... Is my interpretation of calculing the derivative of the definite integral wrong?
 A: Well, in general we can write, if:
$$\text{y}_{\space\text{n}}\left(x\right):=\int_\text{n}^x\text{f}\left(t\right)\space\text{d}t\tag1$$
We get, for the arclength:
$$\mathscr{S}\left(\text{a},\text{b}\right)=\int_\text{a}^\text{b}\sqrt{1+\left(\frac{\text{d}\text{y}_{\space\text{n}}\left(x\right)}{\text{d}x}\right)^2}\space\text{d}x=\int_\text{a}^\text{b}\sqrt{1+\text{f}\left(x\right)^2}\space\text{d}x\tag2$$
So, in this case we get:
$$\mathscr{S}\left(1,4\right)=\int_1^4\sqrt{1+\left(\sqrt{25x^3-1}\right)^2}\space\text{d}x=\int_1^4\sqrt{1+25x^3-1}\space\text{d}x=$$
$$\int_1^4\sqrt{25x^3}\space\text{d}x=\int_1^4\sqrt{25}\cdot\sqrt{x^3}\space\text{d}x=5\cdot\int_1^4\left(x^3\right)^\frac{1}{2}\space\text{d}x=5\cdot\int_1^4x^\frac{3}{2}\space\text{d}x\tag3$$
Now, use:
$$\int x^\text{d}\space\text{d}x=\frac{x^{1+\text{d}}}{1+\text{d}}+\text{C}\tag4$$
So, we get:
$$\mathscr{S}\left(1,4\right)=5\cdot\left[\frac{x^{1+\frac{3}{2}}}{1+\frac{3}{2}}\right]_1^4=5\cdot\left(\frac{4^{1+\frac{3}{2}}}{1+\frac{3}{2}}-\frac{1^{1+\frac{3}{2}}}{1+\frac{3}{2}}\right)=62\tag5$$

In the ultimate general case:
$$\mathscr{S}_{\space\text{n}_1,\text{n}_2}\left(\text{a},\text{b}\right)=\int_\text{a}^\text{b}\sqrt{1+\left(\sqrt{\text{n}_1\cdot x^{\text{n}_2}-1}\right)^2}\space\text{d}x=\int_\text{a}^\text{b}\sqrt{1+\text{n}_1\cdot x^{\text{n}_2}-1}\space\text{d}x=$$
$$\int_\text{a}^\text{b}\sqrt{\text{n}_1\cdot x^{\text{n}_2}}\space\text{d}x=\int_\text{a}^\text{b}\sqrt{\text{n}_1}\cdot\sqrt{x^{\text{n}_2}}\space\text{d}x=\sqrt{\text{n}_1}\cdot\int_\text{a}^\text{b}\left(x^{\text{n}_2}\right)^\frac{1}{2}\space\text{d}x=$$
$$\sqrt{\text{n}_1}\cdot\int_\text{a}^\text{b}x^\frac{\text{n}_2}{2}\space\text{d}x=\sqrt{\text{n}_1}\cdot\left[\frac{x^{1+\frac{\text{n}_2}{2}}}{1+\frac{\text{n}_2}{2}}\right]_\text{a}^\text{b}=\frac{\sqrt{\text{n}_1}}{1+\frac{\text{n}_2}{2}}\cdot\left(\text{b}^{1+\frac{\text{n}_2}{2}}-\text{a}^{1+\frac{\text{n}_2}{2}}\right)\tag6$$
