Calculating the following Integral Here is the question which I need help to calculate:

Knowing that $\displaystyle \int_2^3 (2f(x) - 3g(x)) dx = 2 $   and $\displaystyle \int_2^3 g(t)dt = -1$, 
  find the value of $$\int_2^3 (3f(s) + 2g(s)) ds$$

I do not even know where to start for such a question, any tips of where to start off would be greatly appreciated :)
 A: If
$$\int_{2}^3g(t)dt=-1$$
and
$$\int_{2}^3(2f(x)-3g(x))dx=2$$
then
$$\int_{2}^3(2f(x)-3g(x))dx=2\int_{2}^3f(x)dx-3\int_{2}^3g(x))dx=$$
$$=2\int_{2}^3f(x)dx-3\cdot(-1)=2\Rightarrow\int_{2}^3f(x)dx=-\frac{1}{2}$$
$$\int_{2}^3(3f(s)+2g(s))ds=3\cdot(-\frac{1}{2})+2\cdot(-1)=-\frac{7}{2}$$
A: Setting $A=\int_{2}^{3}f(x)dx, B=\int_{2}^{3}g(x)dx$ may make it easier to solve the question. Then you'll have $$2A-3B=2, B=-1\Rightarrow A=-1/2,B=-1.$$
Hence, the answer will be $$3A+2B=-3/2-2=-7/2.$$
A: This is really essentially an algebra problem, once you use the given conditions and the linearity of the integral. This is one way to proceed: As $\int \limits_{x=2} ^{x=3} 2f - 3g = 2$, and $\int \limits_{x=2} ^{x=3} g = -1$, then
$$
\int \limits_{x=2} ^{x=3} f = 
\frac{1}{2} \int \limits_{x=2} ^{x=3} 2f =
\frac{1}{2} \int \limits_{x=2} ^{x=3} (2f - 3g + 3g) =
\frac{1}{2} \int \limits_{x=2} ^{x=3} (2f - 3g) + \frac{3}{2} \int \limits_{x=2} ^{x=3} g =
\frac{1}{2} \cdot 2 + \frac{3}{2} \cdot (-1) = - \frac{1}{2}
.
$$
Hence, 
$$
\int \limits_{x=2} ^{x=3} 3f + 2g 
= 3 \int \limits_{x=2} ^{x=3} f + 2 \int \limits_{x=2} g 
= - \frac{3}{2} + (-2) = - \frac{7}{2}.
$$
I see others beat me here - however, as I have it written up, perhaps a multitutde of solutions will be helpful, so I will post it anyways.
