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The image contains a mathematical expression written vertically. Let me extract and transcribe the expression:

\[
\int_0^2 \left( 5x - 6 \right)^2 dx
\]

This is a definite integral of the function \( (5x - 6)^2 \) with respect to \( x \), evaluated from 0 to 2.

### Steps to solve the integral:

1. **Expand** \( (5x - 6)^2 \):
   \[
   (5x - 6)^2 = (5x)^2 - 2(5x)(6) + (-6)^2 = 25x^2 - 60x + 36
   \]

2. **Rewrite the integral**:
   \[
   \int_0^2 (25x^2 - 60x + 36) \, dx
   \]

3. **Integrate each term**:
   \[
   \int 25x^2 \, dx = \frac{25x^3}{3}
   \]
   \[
   \int -60x \, dx = -30x^2
   \]
   \[
   \int 36 \, dx = 36x
   \]

4. **Combine the integrals**:
   \[
   \int_0^2 (25x^2 - 60x + 36) \, dx = \left[ \frac{25x^3}{3} - 30x^2 + 36x \right]_0^2
   \]

5. **Evaluate the result at the bounds**:
   First, evaluate at \( x = 2 \):
   \[
   \frac{25(2)^3}{3} - 30(2)^2 + 36(2) = \frac{25(8)}{3} - 30(4) + 72 = \frac{200}{3} - 120 + 72 = \frac{200}{3} - 48 = \frac{56}{3}
   \]

   Then, evaluate at \( x = 0 \) (which gives 0):
   \[
   \frac{25(0)^3}{3} - 30(0)^2 + 36(0) = 0
   \]

6. **Final answer**:
   \[
   \frac{56}{3} \approx 18.67
   \]

Would you like further details or clarifications on this?

### Related questions:
1. How do you apply the power rule for integration?
2. What are the rules for evaluating definite integrals?
3. How can integrals be used to find areas under curves?
4. How does expanding polynomials help simplify integration?
5. What are some common techniques for solving more complex integrals?

**Tip:** Always expand and simplify the integrand when you encounter polynomials before integrating to make the process easier.

Evaluate the definite integral of the function (5x - 6)^2 from 0 to 2.

Learn how to evaluate the definite integral of the function (5x - 6)^2 from 0 to 2. This problem involves polynomial expansion and integration techniques, suitable for students in Grades 11-12.

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