Suppose f is such that f(-x) = -f(x) for ever | vedclass

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(D) Given that $f(-x) = -f(x)$,the function $f$ is an odd function.For an odd function,the property of definite integrals states that $\int_{-a}^{a} f

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$-5$

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(D) Given that $f(-x) = -f(x)$,the function $f$ is an odd function.For an odd function,the property of definite integrals states that $\int_{-a}^{a} f(x) dx = 0$.We can split the integral as: $\int_{-1}^{1} f(x) dx = \int_{-1}^{0} f(x) dx + \int_{0}^{1} f(x) dx = 0$.Substituting the given value $\int_{0}^{1} f(x) dx = 5$,we get: $\int_{-1}^{0} f(x) dx + 5 = 0$.Therefore,$\int_{-1}^{0} f(x) dx = -5$.Since the variable of integration is a dummy variable,$\int_{-1}^{0} f(t) dt = \int_{-1}^{0} f(x) dx = -5$.

Suppose $f$ is such that $f(-x) = -f(x)$ for every real $x$ and $\int_{0}^{1} f(x) dx = 5$,then $\int_{-1}^{0} f(t) dt = $

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