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.
f(29) = f(1) + integral [1, 6] f'(x)dx = 15 + 29 = 44.
To find f(6), we need to integrate f'(t) from 1 to 6. But before that, let's analyze the information given.
We know that f(1) = 15, which tells us the value of f at t = 1.
We also know that f' is continuous, which means that f'(t) is defined and continuous for all values of t in the interval [1, 6]. This assumption allows us to use the Fundamental Theorem of Calculus, which states that the integral of the derivative of a function f'(t) over an interval [a, b] is equal to f(b) - f(a). In other words, the integral of f'(t) from a to b gives us the difference between the values of the function f at points b and a.
So, using the given information, we have:
∫[1 to 6] f'(t) dt = f(6) - f(1) = 29
Now, let's solve for f(6):
f(6) - f(1) = 29
f(6) = 29 + f(1) (adding f(1) to both sides)
f(6) = 29 + 15 = 44
Therefore, f(6) = 44.