True or False
If the function fff is odd, then ∫−11f(x)dx=0\displaystyle{\int_{-1}^{1}f(x)dx=0}∫−11f(x)dx=0
The solution is TrueFrom ∫−11f(x)dx=0.If we let x=−t ⟹ dx=−dt⇒∫−11f(x)dx=∫1−1f(−t)dt=∫−11f(−x)dxIf f is , that is f(−x)=−f(x)⇒∫−11f(x)dx+∫−11f(−x)dx=0Which is True\text{The solution is } \textit{True} \\ \text{From } \int\limits_{ - 1}^1 {f\left( x \right)} dx = 0. \text{If we let } x = - t \implies dx = - dt \\ \Rightarrow \int\limits_{ - 1}^1 {f\left( x \right)} dx = \int\limits_1^{ - 1} {f\left( { - t} \right)} dt = \int\limits_{ - 1}^1 {f\left( { - x} \right)} dx \\ \text{If } f \text{ is , that is } f\left( { - x} \right) = - f\left( x \right) \\ \Rightarrow \int\limits_{ - 1}^1 {f\left( x \right)} dx + \int\limits_{ - 1}^1 {f\left( { - x} \right)} dx = 0 \\ \text{Which is } \textbf{\textit{True}}The solution is TrueFrom −1∫1f(x)dx=0.If we let x=−t⟹dx=−dt⇒−1∫1f(x)dx=1∫−1f(−t)dt=−1∫1f(−x)dxIf f is , that is f(−x)=−f(x)⇒−1∫1f(x)dx+−1∫1f(−x)dx=0Which is True
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