Evaluate the definite integral int_{1}^{2} (4 | vedclass

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(D) Let $I = \int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) dx$. First,find the indefinite integral: $\int (4x^{3} - 5x^{2} + 6x + 9) dx = 4(\frac{x^{4}}{4})

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(D) Let $I = \int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) dx$.
First,find the indefinite integral:
$\int (4x^{3} - 5x^{2} + 6x + 9) dx = 4(\frac{x^{4}}{4}) - 5(\frac{x^{3}}{3}) + 6(\frac{x^{2}}{2}) + 9x = x^{4} - \frac{5x^{3}}{3} + 3x^{2} + 9x = F(x)$.
By the second fundamental theorem of calculus,$I = F(2) - F(1)$.
$F(2) = (2)^{4} - \frac{5(2)^{3}}{3} + 3(2)^{2} + 9(2) = 16 - \frac{40}{3} + 12 + 18 = 46 - \frac{40}{3} = \frac{138 - 40}{3} = \frac{98}{3}$.
$F(1) = (1)^{4} - \frac{5(1)^{3}}{3} + 3(1)^{2} + 9(1) = 1 - \frac{5}{3} + 3 + 9 = 13 - \frac{5}{3} = \frac{39 - 5}{3} = \frac{34}{3}$.
Therefore,$I = \frac{98}{3} - \frac{34}{3} = \frac{64}{3}$.

Evaluate the definite integral $\int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) dx$.

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