If int_{n}^{n+1} g(x) dx = n^2, forall n in Z | vedclass

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(A) Given,$\int_n^{n+1} g(x) dx = n^2, \forall n \in Z$. We need to evaluate $\int_{-3}^3 g(x) dx$. Using the additive property of definite integrals,

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$19$

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(A) Given,$\int_n^{n+1} g(x) dx = n^2, \forall n \in Z$. We need to evaluate $\int_{-3}^3 g(x) dx$. Using the additive property of definite integrals,we can write: $\int_{-3}^3 g(x) dx = \int_{-3}^{-2} g(x) dx + \int_{-2}^{-1} g(x) dx + \int_{-1}^0 g(x) dx + \int_0^1 g(x) dx + \int_1^2 g(x) dx + \int_2^3 g(x) dx$. Substituting the given values for each interval: For $n = -3: \int_{-3}^{-2} g(x) dx = (-3)^2 = 9$. For $n = -2: \int_{-2}^{-1} g(x) dx = (-2)^2 = 4$. For $n = -1: \int_{-1}^0 g(x) dx = (-1)^2 = 1$. For $n = 0: \int_0^1 g(x) dx = (0)^2 = 0$. For $n = 1: \int_1^2 g(x) dx = (1)^2 = 1$. For $n = 2: \int_2^3 g(x) dx = (2)^2 = 4$. Summing these values: $\int_{-3}^3 g(x) dx = 9 + 4 + 1 + 0 + 1 + 4 = 19$.

If $\int_{n}^{n+1} g(x) dx = n^2, \forall n \in Z$,then the value of $\int_{-3}^3 g(x) dx$ is

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