The given integral is:

$\int_0^2 \left(2x + 5\right) \, \mathrm{d}x$

Step-by-step Solution:

1. Break up the integral: The integrand can be split into two simpler integrals: $\int_0^2 (2x + 5) \, \mathrm{d}x = \int_0^2 2x \, \mathrm{d}x + \int_0^2 5 \, \mathrm{d}x$

2. Solve each integral separately:

   • For the first part, $\int_0^2 2x \, \mathrm{d}x$: $\int 2x \, \mathrm{d}x = x^2 + C$ Now, evaluate this from 0 to 2: $\left[ x^2 \right]_0^2 = (2^2) - (0^2) = 4$

   • For the second part, $\int_0^2 5 \, \mathrm{d}x$: $\int 5 \, \mathrm{d}x = 5x + C$ Now, evaluate this from 0 to 2: $\left[ 5x \right]_0^2 = 5(2) - 5(0) = 10$

3. Add the results: $\int_0^2 (2x + 5) \, \mathrm{d}x = 4 + 10 = 14$

Final Answer:

$\int_0^2 (2x + 5) \, \mathrm{d}x = 14$

Would you like more details or have any questions?

Here are some related questions you might want to explore:

1. What is the definite integral of a quadratic function like $\int_0^2 (x^2 + x) \, \mathrm{d}x$?
2. How do you integrate a function with limits that involve infinity?
3. What is the significance of the Fundamental Theorem of Calculus in evaluating definite integrals?
4. How would the result change if the limits of integration were from 1 to 3 instead of 0 to 2?
5. Can you explain how the integral $\int \frac{1}{x} \, \mathrm{d}x$ behaves differently from polynomial integrals?

Tip: When integrating polynomials, each term can be integrated separately, and constants can be factored out of the integral for simpler calculations.