I'm given: $$\int_{5}^{-2} [3f(x) + 1]\,dx$$
with the additional information that: $$\int_{0}^{5} f(x)\,dx = 10$$
and $$\int_{0}^{-2} f(x)\,dx = -4$$
My layman mind looks at it as, since the sum of the two function pieces = 6, then the integral is $3(6) + 1 == 19$
Is that the right way of looking at the problem, or am I missing something?
Thanks for any assistance.