Completely stumped by this supposedly (easy?) calculus question...?

Question

I suddenly ran into a problem I've never seen before...a variation I suppose? I have no idea how to work this one out, but some other guys say they found a solution, but right now, I'm thinking no solution...here's the question:

BTW: ~ is the integration sign (sorry I don't know how to do it on comp) so 1~2 would mean integrating from 1 to 2.

Let:
1~2 f(x) dx = -3,
1~5 f(x) dx = 5, and
1~5 g(x) dx = 9.

Find: 1~2 (3 f(x) +1) dx.

I've done tons of just integration with these kinds of problems, but I've never seen one where there is a constant inside...how can you solve this? Others say the answer is -8.

Thank you everyone

Answer 1

The Sum and Differences Rule of Integration states that

⌠b -------------------- ⌠b --------- ⌠b
⌡a [f(x) ± g(x)]dx = ⌡a f(x)dx ± ⌡a g(x)dx

That means we can separate 3f(x) from 1.

⌠2 ------------⌠2
⌡1 3f(x)dx + ⌡1 (1)dx

We already given the integral of f(x) from x = 1 to x = 2, which is -3.

3(-3) + [(2) - (1)] = -9 + 1 = -8

* Note * Ignore the dashed lines. It's the only way I could maintain the formatting.

Answer 2

You don't need everything there.

[From x = 1 to 2] ∫f(x)dx = F(2) - F(1) = -3

[From x = 1 to 2] ∫{3f(x) + 1}dx = 3[F(2) - F(1)] + x [from 1 to 2]
= 3(-3) + (2 - 1) = -9 + 1 = -8

Answer 3

The way you do this is to simple transform the integral into 1-2 3f(x) + 1-2 1, the integral of y=1 from one to two is the area of a rectangle (this area is one). Just add one to three times the integral of f(x) from one to three, this gives you -9 + 1. The answer should be -8.

Answer 4

1~2 f(x)dx=F(2)-F(1)= - 3


1~2 (3f(x)+1)dx=

=3 (1~2 f(x) dx) + 1~2 (1) =

=3 (F(2) - F(1)) + 2-1=

=(3* -3)+1= - 8