Solution
As I correctly understand the formula in the question is ∫0xf(t)dt=2cosx+3x−2 .
If F(x)=∫0xf(t)dt then f(x) = F’(x).
Therefore f(x) = d(2cosx+3x-2)/dx = -2sinx+3.
Let's check
∫0xf(t)dt=∫0x(−2sint+3)dt=(2cost+3t)∣0x=2cosx+3x−2cos0=2cosx+3x−2
Answer
f(x) = -2sinx+3
Comments