Definite integral explanation of logic If $$\int_{-1}^{4} f(x)dx=4$$ and $$\int_{2}^{4} (3-f(x))dx=7$$, then what will be the value of $$\int_{2}^{-1} f(x)dx?$$ I need to know the logic behind solving these kind of problems. 
 A: This is an exercise in applying the standard properties of the Riemann integral.
We have that
$$
4 = \int_{-1}^4 f(x) dx = - \int_{4}^{-1} f(x) dx \implies \int_{4}^{-1} f(x) dx = -4.
$$
Also
$$
7 = \int_{2}^{4} (3-f(x))dx = \int_{2}^{4} 3 dx - \int_{2}^4 f(x) dx = 6 - \int_{2}^4 f(x) dx.
$$
Hence
$$
\int_{2}^4 f(x) dx = -1.
$$
Finally by above
$$
\int_{2}^{-1} f(x) dx = \int_{2}^4 f(x) dx + \int_{4}^{-1} f(x)dx = (-1)  + (-4) = -5
$$
A: $$\int _{ -1 }^{ 4 } f(x)dx=\int _{ -1 }^{ 2 }{ f\left( x \right) dx+\int _{ 2 }^{ 4 }{ f\left( x \right) dx }  } $$
and we know that $$ \int _{ -1 }^{ 2 }{ f\left( x \right) dx=-\int _{ 2 }^{ -1 }{ f\left( x \right) dx }  } $$
and from here we get $$\int _{ -1 }^{ 4 } f(x)dx=\int _{ 2 }^{ 4 }{ f\left( x \right) dx } -\int _{ 2 }^{ -1 }{ f\left( x \right) dx } $$
now $$\int _{ 2 }^{ 4 } (3-f(x))dx=7\Rightarrow \int _{ 2 }^{ 4 }{ f\left( x \right)  } dx=-1$$ now put all in the expresion
$$4=-1-\int _{ 2 }^{ -1 }{ f\left( x \right) dx } \\$$

$$ \int _{ 2 }^{ -1 }{ f\left( x \right) dx } =\color{red} {-5}$$

