To evaluate the definite integral $\int_0^1 (4x^3 - 2x) \, dx$, we'll proceed step by step:

Step 1: Find the antiderivative

The integral given is:

$\int_0^1 (4x^3 - 2x) \, dx$

We can integrate term by term:

1. The antiderivative of $4x^3$ is $\frac{4x^4}{4} = x^4$.
2. The antiderivative of $-2x$ is $-\frac{2x^2}{2} = -x^2$.

So, the antiderivative of the expression is:

$F(x) = x^4 - x^2$

Step 2: Evaluate the definite integral

Now, we evaluate $F(x)$ at the upper and lower limits:

$\int_0^1 (4x^3 - 2x) \, dx = \left[ x^4 - x^2 \right]_0^1$

Plug in the upper limit $x = 1$:

$F(1) = 1^4 - 1^2 = 1 - 1 = 0$

Plug in the lower limit $x = 0$:

$F(0) = 0^4 - 0^2 = 0$

Thus:

$\int_0^1 (4x^3 - 2x) \, dx = 0 - 0 = 0$

Final Answer:

$\int_0^1 (4x^3 - 2x) \, dx = 0$

Would you like to explore more details about definite integrals or have any questions?

Here are 5 related questions:

1. What is the fundamental theorem of calculus and how does it relate to definite integrals?
2. How do you find the antiderivative of a polynomial function?
3. What are the properties of definite integrals when the result is zero?
4. How do you evaluate definite integrals involving trigonometric functions?
5. What changes when evaluating the integral over a different interval, like $[0, 2]$?

Tip: Always check if the function you're integrating is odd or even, as this can sometimes simplify the computation of definite integrals, especially over symmetric intervals like $[-a, a]$.