To solve the definite integral:

$\int_{2}^{4}(-x^2 + 6x - 8) \, dx$

we will follow these steps:

Step 1: Find the indefinite integral

First, integrate the function term by term:

$\int (-x^2 + 6x - 8) \, dx$

• For $-x^2$, the integral is: $\int -x^2 \, dx = -\frac{x^3}{3}$

• For $6x$, the integral is: $\int 6x \, dx = 3x^2$

• For $-8$, the integral is: $\int -8 \, dx = -8x$

So, the indefinite integral is: $-\frac{x^3}{3} + 3x^2 - 8x + C$

Step 2: Apply the limits of integration

Now, apply the limits of integration from $x = 2$ to $x = 4$:

$\int_{2}^{4} (-x^2 + 6x - 8) \, dx = \left[ -\frac{x^3}{3} + 3x^2 - 8x \right]_{2}^{4}$

This means we need to evaluate the expression at $x = 4$ and $x = 2$ and subtract:

1. At $x = 4$: $-\frac{(4)^3}{3} + 3(4)^2 - 8(4) = -\frac{64}{3} + 3(16) - 32 = -\frac{64}{3} + 48 - 32 = -\frac{64}{3} + 16$ $= \frac{-64 + 48}{3} = \frac{-16}{3}$

2. At $x = 2$: $-\frac{(2)^3}{3} + 3(2)^2 - 8(2) = -\frac{8}{3} + 3(4) - 16 = -\frac{8}{3} + 12 - 16 = -\frac{8}{3} - 4$ $= \frac{-8 - 12}{3} = \frac{-20}{3}$

Step 3: Subtract the results

Now subtract the result at $x = 2$ from the result at $x = 4$:

$\left( \frac{-16}{3} \right) - \left( \frac{-20}{3} \right) = \frac{-16 + 20}{3} = \frac{4}{3}$

Final Answer:

$\int_{2}^{4} (-x^2 + 6x - 8) \, dx = \frac{4}{3}$

Would you like a further breakdown or have any questions on this?

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Here are some related questions for deeper understanding:

1. How do you verify a definite integral using geometry?
2. What is the geometric interpretation of this integral on the coordinate plane?
3. How would you find the indefinite integral of a more complex polynomial?
4. How do you apply the Fundamental Theorem of Calculus to definite integrals?
5. Can this process be automated using numerical integration techniques?

Tip: Always double-check your integration by differentiating the result to ensure it matches the original function.