let (sigma notation) [5, 14] f(x)dx=1, [5,8] f(x)dx=10, [11,14] f(x)=8
Find [8, 11] f(x)dx
and [11, 8] f(x)dx
We went over the rules of computing integrals in class, but I don't see how that helps with this problem.
let (sigma notation) [5, 14] f(x)dx=1, [5,8] f(x)dx=10, [11,14] f(x)=8
Find [8, 11] f(x)dx
and [11, 8] f(x)dx
We went over the rules of computing integrals in class, but I don't see how that helps with this problem.
Hey rhcpule3.
For your problem you have conflicting descriptions of your interval: Do you mean that [5,14] \ [5,8] OR [11,14] has f(x)dx = 1 (in other words any part of [5,14] that is also not in [5,8] or [11,14] has f(x)dx = 1)?
With that "dx" I think you must mean an integral rather than a sum (sigma).
You are given that $\displaystyle \int_5^{14} f(x)dx= 1$, that $\displaystyle \int_5^{8} f(x)dx= 10$, that $\displaystyle \int_{11}^{14} f(x)= 8$.
You want to find $\displaystyle \int_8^{11} f(x)dx$ and $\displaystyle \int_{11}^8 f(x)dx$.
One of the "rules of computing intervals" is that $\displaystyle \int_a^b f(x) dx+ \int_b^c f(x)dx= \int_a^c f(x)dx$ which means that
$\displaystyle \int_5^8 f(x)dx+ \int_8^{11} f(x)dx= \int_5^11 f(x)dx$ and
$\displaystyle \int_5^{11} f(x)dx+ \int_{11}^{14} f(x)dx= \int_5^{14} f(x)dx$.
Putting in the information given those become
$\displaystyle 10+ \int_8^{11} f(x) dx= \int_5^{11} f(x)dx$ and
$\displaystyle \int_5^{11} f(x)dx+ 8= 1$
Can you solve 10+ x= y and y+ 8= 1 for x?