To evaluate the definite integral

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

we'll follow these steps:

1. Find the indefinite integral of $x^2 - 2x - 3$.
2. Evaluate the antiderivative at the upper and lower limits of 8 and 6.
3. Subtract the results to find the final answer.

Step 1: Integrate $x^2 - 2x - 3$

The antiderivative of each term is:

• $\int x^2 \, dx = \frac{x^3}{3}$,
• $\int -2x \, dx = -x^2$,
• $\int -3 \, dx = -3x$.

So, the indefinite integral is:

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

Step 2: Apply the Limits

Now, we evaluate this at $x = 8$ and $x = 6$:

$\left( \frac{8^3}{3} - 8^2 - 3 \cdot 8 \right) - \left( \frac{6^3}{3} - 6^2 - 3 \cdot 6 \right).$

Calculating each part:

For $x = 8$: $\frac{8^3}{3} = \frac{512}{3}, \quad 8^2 = 64, \quad 3 \cdot 8 = 24.$ So, $\frac{512}{3} - 64 - 24 = \frac{512 - 264}{3} = \frac{248}{3}.$

For $x = 6$: $\frac{6^3}{3} = \frac{216}{3} = 72, \quad 6^2 = 36, \quad 3 \cdot 6 = 18.$ So, $72 - 36 - 18 = 18.$

Step 3: Final Calculation

Now subtract the values: $\frac{248}{3} - 18 = \frac{248 - 54}{3} = \frac{194}{3}.$

Final Answer:

$\int_{6}^{8} (x^2 - 2x - 3) \, dx = \frac{194}{3}.$

Would you like more details or have any questions?

Related Questions:

1. How would you evaluate an integral with different polynomial terms?
2. How can integration be applied in real-world contexts?
3. What is the importance of using limits in definite integrals?
4. How would changing the limits affect the integral result?
5. How does the Fundamental Theorem of Calculus apply to this problem?

Tip: Always verify each step when calculating definite integrals to avoid arithmetic errors.