If ∫31(6f(x)+5)dx=13, then
∫^3_1f(x)dx=
We can use linear properties of integrals to solve this problem.
First, we can distribute the constant 6 from the integrand:
∫3^1 (6f(x) + 5) dx = 6∫3^1 f(x) dx + ∫3^1 5 dx
Next, we can use the fact that ∫3^1 5 dx is the area of a rectangle with height 5 and base of length 2, which is simply 10.
So we have:
∫3^1 (6f(x) + 5) dx = 6∫3^1 f(x) dx + 10
But we are given that ∫3^1 (6f(x) + 5) dx = 13, so we can substitute:
13 = 6∫3^1 f(x) dx + 10
Solving for ∫3^1 f(x) dx, we find:
∫3^1 f(x) dx = (13 - 10)/6 = 1/2
Therefore, the answer is 1/2.