The integral shown in the image is:

$\int_0^1 3x^2 \, dx$

To solve this, follow these steps:

1. First, find the antiderivative of $3x^2$. The antiderivative of $x^n$ is $\frac{x^{n+1}}{n+1}$, so:

$\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$

2. Now, apply the limits of integration (from 0 to 1):

$\left[ x^3 \right]_0^1 = 1^3 - 0^3 = 1 - 0 = 1$

So, the value of the integral is:

$1$

Let me know if you'd like more details or have any questions! Here are five related questions to explore:

1. What is the integral of $x^3$ over the same interval?
2. How would the result change if the limits were from 1 to 2 instead?
3. What is the geometric interpretation of this integral?
4. How do you integrate higher powers of $x$, such as $x^4$?
5. How does changing the coefficient (like from 3 to 5) affect the integral?

Tip: When integrating powers of $x$, always increase the exponent by 1 and divide by the new exponent!