If f(1) = 15
f' is continuous
and integral from 1 to 6 of f'(t)dt = 29
what is f(6)?
I took the antiderivative as F(6)-F(1) so it's 6x-1x and set it to 29... but that really does not do anything, especially if you plug in 1.
1 answer
f(29) = f(1) + integral [1, 6] f'(x)dx = 15 + 29 = 44.