Question

Find the area of the region bounded by the graphs of the equations

Answer 1

Ok, so this is integration between two regions, do you know the formula?

Answer 2

\[A = \int\limits_{b}^{a}[f(x)-g(x)]dx\]

Answer 3

ok so
f(x)=sin(x)
g(x)=cos(2x)
and the interval is
\[\Huge \left[ -\frac{\pi}{2},\frac{\pi}{6} \right]\]

Answer 4

\[\Large A = \int\limits\limits_{-\frac{\pi}{2}}^{\frac{\pi}{6}}[\sin(x)-\cos(2x)]dx\]

Answer 5

\[\Large = \int\limits\limits\limits_{-\frac{\pi}{2}}^{\frac{\pi}{6}}[\sin(x)]dx-\int\limits\limits\limits_{-\frac{\pi}{2}}^{\frac{\pi}{6}}[\cos(2x)]dx\]

Answer 6

Basically find the two seperate defintie integrals and hten you subtract the answers

Answer 7

I get -2.11272 with that. I already got that, but it doesn't match the correct answer. I was told by a classmate the answer was E. E = 1.29904

Answer 8

Ok well, I put it in a calculator and got E as the answer so lets work through it.

Answer 9

The integrals are -cos(x) and 0.5sin(2x) respectively for sin(x) and cos(2x)

Answer 10

NVM, my calculator was on degree mode. I get -1.29904. My question is why does my answer come out negative and the answer is positive?

Answer 11

I was wondering the same thing for a second, but look at the key word: AREA, they want to know how much space is inbetween, sure part of it might be underneath the x-axis but this is almost like them asking how much space is between two gardens? You can say -2m^s, that doesn't make sense, this is asking for area BETWEEN two things so you have to treat it like a real world question, do you get it?

Answer 12

Yes I do, thanks a lot