The given expression is a definite integral:

$\int_1^4 \left(x^3 - 3x^2 + 4x \right) \, dx$

To solve this, we will integrate the function $f(x) = x^3 - 3x^2 + 4x$ and then evaluate it at the upper and lower limits (4 and 1).

Step-by-Step Solution:

1. Integrate each term of the function:

   • For $x^3$, the integral is $\frac{x^4}{4}$.
   • For $-3x^2$, the integral is $-x^3$.
   • For $4x$, the integral is $2x^2$.

2. The indefinite integral becomes:

$\int (x^3 - 3x^2 + 4x) \, dx = \frac{x^4}{4} - x^3 + 2x^2 + C$

3. Apply the limits of integration (from 1 to 4):

$\left[ \frac{x^4}{4} - x^3 + 2x^2 \right]_1^4$

4. Evaluate at $x = 4$:

$\frac{4^4}{4} - 4^3 + 2(4^2) = \frac{256}{4} - 64 + 2(16) = 64 - 64 + 32 = 32$

5. Evaluate at $x = 1$:

$\frac{1^4}{4} - 1^3 + 2(1^2) = \frac{1}{4} - 1 + 2 = \frac{1}{4} + 1 = \frac{5}{4}$

6. Subtract the results:

$32 - \frac{5}{4} = \frac{128}{4} - \frac{5}{4} = \frac{123}{4}$

Thus, the value of the integral is:

$\frac{123}{4}$

Would you like further details on the solution or have any other questions?

Here are some related questions you might find interesting:

1. How do you evaluate integrals with different powers of $x$?
2. What are the rules for integrating polynomials?
3. Can you explain how the fundamental theorem of calculus is used here?
4. How do definite integrals relate to the area under a curve?
5. What is the difference between an indefinite and a definite integral?

Tip: Always remember to check the integral of each term separately when handling polynomials in an integral.