Question: I am trying to solve this task 

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> Consider the measure space $([0,1],\mathcal{B}([0,1]),\mu:=2\lambda\_1+3\delta\_{1})$ and the measurable function $f(\omega)=2-\omega$. Find $$\int\_{[0,1]}f\,d\mu$$
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My Solution

$$\int\_{a}^{b} f(x) d x=\int\_{[a, b]} f d \lambda\_{1}$$

We know

Thus

$$\int\_{[0,1]} f d \mu=\int\_{[0,1]} 2- x\left(2d \lambda\_{1}+3 d \delta\_{1}\right)=\cdots=6$$

Answer: 
I pressume that $\delta\_1$ is the Dirac measure and $\lambda\_1$ the Lebesgue measure.

If so,then you are correct.

Indeed:

So $$\int\_0^1f(x)d\mu(x)=2\int\_0^1(2-x)d\lambda\_1(x)+3\int\_0^1f(x)d\delta\_1(x)$$ $$=2\int\_0^1(2-x)dx+3 f(1)=2(2-\frac{1}{2})+3=6$$
Marios answer is correct

Just wanted to note to the OP some general facts. It can be checked for simple functions, and then proven in general, that for $a,b$ non-negative and $\mu = a\lambda + b\delta$ where $\lambda,\delta$ finite on the measurable set $A$, we have
$$ \int\_{A} f d\mu = a\int\_{A} f d\lambda + b\int\_{A} f d\delta $$