<s>[INST] 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 :)
 [/INST] 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.
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