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

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(B) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$ $= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$

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$\frac{7}{3}$

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(B) First, we evaluate the definite integral: $\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$ $= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$. Thus, $k = 15$. Now, we solve the equation $3x^2 - 4x + 2 = \frac{3k}{5}$: $3x^2 - 4x + 2 = \frac{3(15)}{5} = 9$. $3x^2 - 4x - 7 = 0$. Factoring the quadratic equation: $3x^2 - 7x + 3x - 7 = 0$ $x(3x - 7) + 1(3x - 7) = 0$ $(x + 1)(3x - 7) = 0$. The roots are $x = -1$ and $x = \frac{7}{3}$. Since we require the root in the interval $[0, 3]$, the valid root is $x = \frac{7}{3}$.

If $\int_0^3 (3x^2 - 4x + 2) dx = k$, then a root of $3x^2 - 4x + 2 = \frac{3k}{5}$ that lies in the interval $[0, 3]$ is

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