To evaluate the definite integral

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx,$

we follow these steps:

---

Step 1: Find the indefinite integral

First, compute the indefinite integral of the function $4x^2 - 5x + 6$:

$\int \left(4x^2 - 5x + 6\right) dx = \frac{4x^3}{3} - \frac{5x^2}{2} + 6x + C$

---

Step 2: Evaluate the definite integral

Now, evaluate the definite integral by substituting the bounds of integration:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \left[\frac{4x^3}{3} - \frac{5x^2}{2} + 6x\right]_{-3}^{5}.$

Substitute $x = 5$:

$\frac{4(5)^3}{3} - \frac{5(5)^2}{2} + 6(5) = \frac{4(125)}{3} - \frac{5(25)}{2} + 30.$

Simplify each term:

$\frac{4(125)}{3} = \frac{500}{3}, \quad \frac{5(25)}{2} = \frac{125}{2}, \quad 6(5) = 30.$

So, at $x = 5$:

$\frac{500}{3} - \frac{125}{2} + 30.$

Next, substitute $x = -3$:

$\frac{4(-3)^3}{3} - \frac{5(-3)^2}{2} + 6(-3) = \frac{4(-27)}{3} - \frac{5(9)}{2} + (-18).$

Simplify each term:

$\frac{4(-27)}{3} = -36, \quad \frac{5(9)}{2} = \frac{45}{2}, \quad 6(-3) = -18.$

So, at $x = -3$:

$-36 - \frac{45}{2} - 18.$

---

Step 3: Combine results

Now subtract the results at $x = -3$ from the results at $x = 5$:

1. At $x = 5$: $\frac{500}{3} - \frac{125}{2} + 30$
2. At $x = -3$: $-36 - \frac{45}{2} - 18$

Simplify:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \left(\frac{500}{3} - \frac{125}{2} + 30\right) - \left(-36 - \frac{45}{2} - 18\right).$

Expand and combine:

$= \frac{500}{3} - \frac{125}{2} + 30 + 36 + \frac{45}{2} + 18.$

Group fractions and constants:

$= \frac{500}{3} + \left(-\frac{125}{2} + \frac{45}{2}\right) + \left(30 + 36 + 18\right).$

Simplify each group:

1. Constants: $30 + 36 + 18 = 84$,
2. Fractions: $-\frac{125}{2} + \frac{45}{2} = -\frac{80}{2} = -40$.

Combine everything:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \frac{500}{3} - 40 + 84.$

Simplify:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \frac{500}{3} + 44.$

Express as a single fraction:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \frac{500}{3} + \frac{132}{3} = \frac{632}{3}.$

---

Final Answer:

$\int_{-3}^{5} \left(4x^2 - 5x + 6\right) dx = \frac{632}{3}.$

---

Would you like me to explain any step in more detail or provide further examples?

Related Questions:

1. How do definite integrals represent the area under the curve?
2. What is the role of the Fundamental Theorem of Calculus in evaluating integrals?
3. How do you handle improper integrals where the bounds are infinite?
4. Can you explain the geometric interpretation of negative results in definite integrals?
5. How can definite integrals be used in physics or engineering applications?

Tip:

Always simplify terms systematically when evaluating definite integrals to avoid errors in arithmetic!