The value of int_{-3}^{3} (ax^5 + bx^3 + cx | vedclass

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(B) Let $I = \int_{-3}^{3} (ax^5 + bx^3 + cx + k) dx$. We can split the integral as: $I = \int_{-3}^{3} (ax^5 + bx^3 + cx) dx + \int_{-3}^{3} k dx$. S

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$k$

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(B) Let $I = \int_{-3}^{3} (ax^5 + bx^3 + cx + k) dx$. We can split the integral as: $I = \int_{-3}^{3} (ax^5 + bx^3 + cx) dx + \int_{-3}^{3} k dx$. Since $f(x) = ax^5 + bx^3 + cx$ is an odd function (i.e.,$f(-x) = -f(x)$),the integral of this function over the symmetric interval $[-3, 3]$ is $0$. Therefore,$I = 0 + \int_{-3}^{3} k dx = \int_{-3}^{3} k dx$. Evaluating this,we get $I = [kx]_{-3}^{3} = k(3) - k(-3) = 3k + 3k = 6k$. Thus,the value of the integral depends only on the constant $k$.

The value of $ \int_{-3}^{3} (ax^5 + bx^3 + cx + k) dx $,where $a, b, c, k$ are constants,depends only on . . . . . .

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