Let f(x)=∫0x tsin1tdt. Then the number of points of discontinuity of the function f(x) in the open interval (0,π) is
0
1
2
infinite
f(x)=xsin1x. At all points in (0,π),f′(x) has a definite finite value.
∴ f(x) is differentiable finitely in (0,π). As a finitely differentiable
function is also continuous, hence f(x) is continuous in (0,π).