If int_{n}^{n+1} f(x) dx = n^2 + n for all n | vedclass

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(C) Given that $\int_{n}^{n+1} f(x) dx = n^2 + n$. We need to evaluate $\int_{-3}^{3} f(x) dx$. Using the property of definite integrals,we can write:

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$16$

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(C) Given that $\int_{n}^{n+1} f(x) dx = n^2 + n$. We need to evaluate $\int_{-3}^{3} f(x) dx$. Using the property of definite integrals,we can write: $\int_{-3}^{3} f(x) dx = \int_{-3}^{-2} f(x) dx + \int_{-2}^{-1} f(x) dx + \int_{-1}^{0} f(x) dx + \int_{0}^{1} f(x) dx + \int_{1}^{2} f(x) dx + \int_{2}^{3} f(x) dx$. Now,substitute $n = -3, -2, -1, 0, 1, 2$ into the given formula $n^2 + n$: For $n = -3$: $(-3)^2 + (-3) = 9 - 3 = 6$. For $n = -2$: $(-2)^2 + (-2) = 4 - 2 = 2$. For $n = -1$: $(-1)^2 + (-1) = 1 - 1 = 0$. For $n = 0$: $(0)^2 + 0 = 0$. For $n = 1$: $(1)^2 + 1 = 2$. For $n = 2$: $(2)^2 + 2 = 6$. Summing these values: $6 + 2 + 0 + 0 + 2 + 6 = 16$.

If $\int_{n}^{n+1} f(x) dx = n^2 + n$ for all $n \in I$,then the value of $\int_{-3}^{3} f(x) dx$ is equal to

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