If \(f(x+y)=f(x)+f(y)\), then what is \(\displaystyle\int_{-1}^{1}f(x)\,dx\) equal to?
0
Putting \(x=y=0\) gives \(f(0)=0\), and putting \(y=-x\) gives \(f(-x)=-f(x)\), so \(f\) is an odd function. The integral of an odd function over the symmetric interval \([-1,1]\) is always zero, so \(\int_{-1}^{1}f(x)\,dx=0\).
Similar Questions
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The value of \(\int^{\pi / 3}_{-\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx\) is
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