When does a part of an equation come before the integral sign?

Question

I have the following equation:

$$V = \int \pi \ f^2(x) \mathrm dx \text{ where } f(x) = 0.4x$$

My answer was:

1. Integrate $\pi$: $\int \pi =\left(\frac12\right)\pi^2$

2. Integrate $ f^2(x) $: $\int f^2(x)=\left(\frac13\right)(0.4x)^3$

3. Add together: $V = $$\int \pi \ f^2(x) dx = \left(\frac12\right)\pi^2 \cdot \left(\frac13\right)(0.4x)^3$

This was not corrent, the correct answer was: $$V = \int \pi \ f^2(x)\mathrm dx = \int \pi \cdot (0.4x)^2 \mathrm dx = \left(\frac{4 \pi}{25}\right)\int x^2 \mathrm dx = \left(\frac{4 \pi}{25}\right) \left(\frac13\right)x^3$$

So why does $\left(\frac{4 \pi}{25}\right)$ come in front of the integral sign in the third step of the solution?

Answer 1

Because, for any constant $k$, $\int kf(x)\,\mathrm dx=k\int f(x)\,\mathrm dx$.

Answer 2

$\pi$ is not a variable, it is a constant. Therefore, the power rule does not apply here.

$V = \int \pi(\frac{2}{5}x)^2 \mathrm dx = \int (\frac{4\pi }{25}) \ x^2 \mathrm dx$. Since the area under $(\frac{4\pi }{25}) \ x^2 $ is just the area under $x^2$, scaled by $\frac{4\pi }{25}$ in the $y$-direction, we can just take the constant out to get $\frac{4\pi}{25}$ $\int x^2 \mathrm dx$.

Now we can apply the power rule to $\int x^2 \mathrm dx$, to get $\frac{x^3}{3}$. Therefore the integral is $\frac{4\pi}{25} \cdot\frac{x^3}{3} = \frac{4\pi x^3}{75}$.