Let f:R→R be a function defined as f(x)=asin2π[x]+[2−x], a∈R, where [t] is the greatest integer less than or equal to t. If x→−1limf(x) exists, then the value of ∫04f(x)dx is equal to
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Official solution
For the limit to exist at x=−1: x→−1−limf(x)=x→−1+limf(x), i.e. asin(−π)+3=asin(−2π)+2, giving 3=−a+2⟹a=−1.
\[\int_0^4 f(x)\,dx = \int_0^4\left(-\sin\frac{\pi[x]}2+[2-x]\right)dx = \int_0^1 1\,dx+\int_1^2(-1)\,dx+\int_2^3(-1)\,dx+\int_3^4(-1)\,dx\]
=1−1−1−1=−2