Find $\int _2^5f\left(x\right)dx$ given the additional conditions in each case. Of course, as I'm nearing the end of the semester, I got sick and fell a bit behind in my calculus class. I'm really having a tough time with this question, and I was hoping for some help on this. Thank you.
a. $f\left(x\right)$ is odd and $\int _{-2}^5f\left(x\right)dx=8$
I'm not really sure where to go with this one. I know the properties of odd functions but I'm not sure how to use these conditions to find the answer.
b. $f\left(x\right)$ is even, $\int _{-2}^2f\left(x\right)dx=6$, and $\int _{-5}^5f\left(x\right)dx=11$
This is even, so I think $$\int _0^2f\left(x\right)=3$$ and $$\int _0^5f\left(x\right)=5.5$$ so $$\int _2^5f\left(x\right)dx=2.5$$
c. $\int _2^5\left(2f\left(x\right)+4\right)dx=18$
$$=2\int _2^5f\left(x\right)dx+\int _2^54dx$$
I'm not sure where to go from there. I think I can just find $\int _2^54dx$ and subtract it, and divide 2 from both sides, but this seems off to me.
d. $\int _2^43f\left(x\right)dx=12$ and $\int _5^4f\left(x\right)dx=-1$
I can change $$\int _5^4f\left(x\right)dx=-1$$ to $$-\int _4^5f\left(x\right)dx=1$$, then add 12, so $$\int _2^5f\left(x\right)dx=13$$
 A: a)$$\int _{-2}^5f\left(x\right)dx-\int _{-2}^2f\left(x\right)dx=\int _2^5f\left(x\right)dx$$
But $f$ is odd so $\int _{-2}^2f\left(x\right)dx=0$. Finally :$$\int _2^5f\left(x\right)dx=8$$
b) It is true
c) You have a good idea, you should calculate it.
d) It is true
A: I found the answers, thanks to Jennifer!
a. $f\left(x\right)$ is odd and $\int _{-2}^5f\left(x\right)dx=8$
$f\left(x\right)$ is odd, so $\int _{-2}^2f\left(x\right)dx=0$
$$\int _{-2}^5f\left(x\right)dx-\int _{-2}^2f\left(x\right)dx=\int _2^5f\left(x\right)dx$$
$$\int _2^5f\left(x\right)dx=8-0=8$$
b. $f\left(x\right)$ is even, $\int _{-2}^2f\left(x\right)dx=6$, and $\int _{-5}^5f\left(x\right)dx=11$
$f\left(x\right)$ is even, so $\int _0^2f\left(x\right)=3$ and $\int _0^5f\left(x\right)=5.5$
$$\int _0^5f\left(x\right)-\int _0^2f\left(x\right)=\int _2^5f\left(x\right)dx$$
$$\int _2^5f\left(x\right)dx=5.5-3=2.5$$
c. $\int _2^5\left(2f\left(x\right)+4\right)dx=18$
$$2\int _2^5f\left(x\right)dx+\int _2^54dx=18$$
$$\int4dx=4x+C$$
$$\lim _{x\to 2+}\left(4x\right)=8$$
$$\lim _{x\to 5-}\left(4x\right)=20$$
$$2\int _2^5f\left(x\right)dx+(20-8)=18$$
$$2\int _2^5f\left(x\right)dx+12=18$$
$$2\int _2^5f\left(x\right)dx=6$$
$$\int _2^5f\left(x\right)dx=3$$
d. $\int _2^43f\left(x\right)dx=12$ and $\int _5^4f\left(x\right)dx=-1$
$$-\int _4^5f\left(x\right)dx=1$$
$$\int _2^43f\left(x\right)dx+\int _4^5f\left(x\right)dx=\int _2^5f\left(x\right)dx$$
$$\int _2^5f\left(x\right)dx=12+1=13$$
