Evaluate the definite integral int_{-1}^{1}(x | vedclass

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(C) Let $I = \int_{-1}^{1} (x+1) d x$. First,find the indefinite integral: $\int (x+1) d x = \frac{x^2}{2} + x = F(x)$. By the second fundamental theo

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$2$

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(C) Let $I = \int_{-1}^{1} (x+1) d x$. First,find the indefinite integral: $\int (x+1) d x = \frac{x^2}{2} + x = F(x)$. By the second fundamental theorem of calculus,we have: $I = F(1) - F(-1)$. Substitute the limits: $F(1) = \frac{1^2}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}$. $F(-1) = \frac{(-1)^2}{2} + (-1) = \frac{1}{2} - 1 = -\frac{1}{2}$. Calculate the difference: $I = \frac{3}{2} - (-\frac{1}{2}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$.

Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$.

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