If int_0^3 (3x^2 - 4x + 2) dx = k, then an in | vedclass

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(D) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$. Substituting the limits: $(3^3 - 2(3^2) + 2(3)) -

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$-1$

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(D) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$. Substituting the limits: $(3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$. Thus, $k = 15$. Now, substitute $k = 15$ into the equation $3x^2 - 4x + 2 = \frac{3k}{5}$: $3x^2 - 4x + 2 = \frac{3(15)}{5} = 9$. Rearranging the equation: $3x^2 - 4x + 2 - 9 = 0$, which simplifies to $3x^2 - 4x - 7 = 0$. Factoring the quadratic equation: $(x + 1)(3x - 7) = 0$. The roots are $x = -1$ and $x = \frac{7}{3}$. The integer root is $-1$.

If $\int_0^3 (3x^2 - 4x + 2) dx = k$, then an integer root of $3x^2 - 4x + 2 = \frac{3k}{5}$ is

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