To evaluate the integral $\int_1^9 (x^2 - 4x + 9) \, dx$ using the definition of the integral, we first find the antiderivative of the function $f(x) = x^2 - 4x + 9$. Then we apply the Fundamental Theorem of Calculus to calculate the definite integral.

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Step 1: Find the Antiderivative

The given function is: $f(x) = x^2 - 4x + 9$ The antiderivative $F(x)$ is calculated term by term:

1. $\int x^2 \, dx = \frac{x^3}{3}$,
2. $\int -4x \, dx = -2x^2$,
3. $\int 9 \, dx = 9x$.

Thus, the antiderivative is: $F(x) = \frac{x^3}{3} - 2x^2 + 9x.$

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Step 2: Evaluate the Definite Integral

The definite integral is given by: $\int_1^9 (x^2 - 4x + 9) \, dx = F(9) - F(1),$ where $F(x) = \frac{x^3}{3} - 2x^2 + 9x$.

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Step 3: Compute $F(9)$ and $F(1)$

1. Compute $F(9)$: $F(9) = \frac{9^3}{3} - 2(9^2) + 9(9).$ $F(9) = \frac{729}{3} - 2(81) + 81 = 243 - 162 + 81 = 162.$

2. Compute $F(1)$: $F(1) = \frac{1^3}{3} - 2(1^2) + 9(1).$ $F(1) = \frac{1}{3} - 2 + 9 = \frac{1}{3} + 7 = \frac{22}{3}.$

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Step 4: Subtract $F(9)$ and $F(1)$

$\int_1^9 (x^2 - 4x + 9) \, dx = F(9) - F(1) = 162 - \frac{22}{3}.$ Convert $162$ to a fraction with denominator 3: $162 = \frac{486}{3}.$ $F(9) - F(1) = \frac{486}{3} - \frac{22}{3} = \frac{464}{3}.$

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Final Answer:

$\int_1^9 (x^2 - 4x + 9) \, dx = \frac{464}{3}.$

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Would you like further details or assistance with any step?

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Related Questions:

1. How do you verify the antiderivative by differentiating it?
2. What is the geometric interpretation of this integral?
3. How does the Fundamental Theorem of Calculus connect derivatives and integrals?
4. Can this integral be solved using numerical methods instead?
5. How does the choice of bounds affect the result?

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Tip:

When solving definite integrals, always simplify terms carefully after substitution to avoid errors with fractions or signs!