If \(f\) is integrable on \([0, a]\), then the function \(h\) defined on \([0, a]\) as \(h(x)=\ldots . .…
If \(f\) is integrable on \([0, a]\), then the function \(h\) defined on \([0, a]\) as \(h(x)=\ldots . . \forall x \in[0, a]\) is integrable on \([0, a]\)
\(f(a-x)\)
\(f(x-a)\)
\(f(x)\)
\(f(a)\)
Solution
Given, \(f\) is integrable an \([0, a]\)
\(\begin{aligned}
\therefore \quad h(x) & =\int_0^a f(x) d x \\
h(x) & =\int_0^a f(a-x) d x
\end{aligned}\)
Hence, option (a) is correct.