To evaluate the definite integral:

$\int_{0}^{2} (2x - 3)(8x^2 + 4) \, dx,$

we will expand the integrand and compute the integral term by term.

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Step 1: Expand the product

Expand $(2x - 3)(8x^2 + 4)$:

$(2x - 3)(8x^2 + 4) = (2x)(8x^2) + (2x)(4) - (3)(8x^2) - (3)(4),$ $= 16x^3 + 8x - 24x^2 - 12.$

Thus, the integrand becomes: $16x^3 - 24x^2 + 8x - 12.$

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Step 2: Compute the integral term by term

The integral can now be written as: $\int_{0}^{2} (16x^3 - 24x^2 + 8x - 12) \, dx.$

Integrate each term separately:

1. $\int 16x^3 \, dx = \frac{16x^4}{4} = 4x^4$,
2. $\int -24x^2 \, dx = \frac{-24x^3}{3} = -8x^3$,
3. $\int 8x \, dx = \frac{8x^2}{2} = 4x^2$,
4. $\int -12 \, dx = -12x$.

Thus, the integral becomes: $\left[4x^4 - 8x^3 + 4x^2 - 12x \right]_{0}^{2}.$

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Step 3: Evaluate the definite integral

Evaluate at the bounds $x = 2$ and $x = 0$:

1. At $x = 2$: $4(2)^4 - 8(2)^3 + 4(2)^2 - 12(2) = 4(16) - 8(8) + 4(4) - 12(2),$ $= 64 - 64 + 16 - 24 = -8.$

2. At $x = 0$: $4(0)^4 - 8(0)^3 + 4(0)^2 - 12(0) = 0.$

Thus, the definite integral is: $-8 - 0 = -8.$

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Final Answer:

$\int_{0}^{2} (2x - 3)(8x^2 + 4) \, dx = -8.$

Would you like additional explanations or clarifications? 😊

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Related Questions:

1. How do you handle integrals involving trigonometric functions?
2. What happens if the bounds are reversed in definite integrals?
3. Can you explain the difference between definite and indefinite integrals?
4. How do you evaluate an improper integral?
5. What are some applications of definite integrals in real life?

Tip: Always double-check the limits of integration to avoid sign errors in definite integrals!