If $\int_{-2}^{3} [2f(x)+2]\,dx = 18$, and $\int_1^{-2} f(x)\,dx=-8$, then $\int_1^{3} f(x)\,dx$ is equal to what? If $\int_{-2}^{3} [2f(x)+2]\,dx = 18$, and  $\int_1^{-2} f(x)\,dx=-8$, then $\int_1^{3} f(x)\,dx$ is equal to what?
So I've tried the regular sum rules and tried plugging in everything I can for any case of substitution, but I can't find out a correct answer. I thought 8/3=f(x) from the second integral, but that doesn't hold up for the first integral, because f(x) there is equal to 4. No idea where to go.
 A: If $\int_{-2}^{3} [2f(x)+2]\,dx = 18-> \int_{-2}^{3} f(x)dx=4 $, then $\int_1^{3} f(x)\,dx$= $\int_{1}^{-2} f(x)dx$ +  $\int_{-2}^{3} f(x) dx$=$-8+4=-4$
A: Since $$9=\int_{-2}^{3} [f(x)+1]\,dx = \int_{-2}^{3} f(x)\,dx + x\big{|}^{3}_{-2}$$
so $$\int_{-2}^{3} f(x)\,dx =9-5=4$$
Use:
$$\boxed{\int_a^{b} f(x)\,dx =\int_{a}^{c} f(x)\,dx +\int_c^{b} f(x)\,dx}$$
$$\int_1^{3} f(x)\,dx =\int_{-2}^{3} f(x)\,dx +\int_1^{-2} f(x)\,dx=4-8=-4$$
A: $$\int_{-2}^3 [2f(x)+2]dx= 2\int_{-2}^3 f(x)dx+ 2\int_{-2}^3dx = 2\int_{-2}^3 f(x)dx+10= 2 \int_{-2}^1 f(x)dx + 2 \int_1^3 f(x)dx +10=  -2 \int_1^{-2} f(x)dx + 2 \int_1^3 f(x)dx +10=16+2\int_1^3 f(x) dx +10=26+2 \int_1^3 f(x) dx=1
8$$
$$2\int_1^3 f(x) dx=-8$$
$$\int_1^3 f(x) dx=-4$$
A: Let the antiderivative of $f(x)$ be $F(x)$. Then, by the Fundamental Theorem of Calculus, we get:$$9=\int_{-2}^3 [f(x)+1]dx= (F(x)+x)|_{-2}^3= F(3)-F(-2)+5$$
$$-8=\int_{1}^{-2} f(x) dx=F(-2)-F(1)$$
and adding the two, we get $$1=F(3)-F(1)+5\implies-4=F(3)-F(1)=\int_1^3f(x) dx$$
A: $$\int_{-2}^{3} \Big(2f(x)+2\Big)\,dx = 2\int_{-2}^{3} \Big(f(x)+1\Big)\,dx=18$$ $$\implies \int_{-2}^{3} \Big(f(x)+1\Big)\,dx=9$$ $$\implies \int_{-2}^{3}f(x)\,dx=4$$
Also, if $c\in[a,b]$ then
$$\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$$
therefore
$$\int_{-2}^3 f(x)\,dx =\int_{-2}^{1} f(x)\,dx +\int_1^{3} f(x)\,dx$$
$$4=8+\int_1^{3} f(x)\,dx$$
$$\implies \int_1^{3} f(x)\,dx=\boxed{-4}$$
