Question

Integration

Answer 1

I asked this last night as well and I learned about a function where
\[\int\limits_{-k}^{k}f(x)=0\]
But it was only for odd functions?

Answer 2

You only need to use a couple of things here to make this easy.

Even functions are symmetrical over the y-axis. (ie the integral from 0 to 5, is equal to the integral from 0 to -5)

For an odd function: the integral from 0 to 5, is equal to negative of the integral from 0 to -5.

The average value of a function is equal to \[\large \frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x) dx\]

Answer 3

right, I;ve got all that

Answer 4

\[\frac{ 1 }{ b-a } \int\limits_{a}^{b} f(x)~dx = average~value\]

yeah like since you know that the average value for the function

Answer 5

So what you have here, really, is that you've halved the amount of total area, right? (by making the interval -5 to 0, instead of -5 to 5)

But you've also made the interval half as long.

Answer 6

Right

Answer 7

You're ~given this in the question\[\large \frac{ 1 }{ 10}\int\limits\limits_{-5}^{5}f(x) dx = 4\] which means
\[\large \int\limits\limits\limits_{-5}^{5}f(x) dx = 40\]

Answer 8

oh duh, it all makes sense now

Answer 9

You need\[\large \frac{ 1 }{ 5}\int\limits\limits\limits_{-5}^{0}f(x) dx = ?\]
But you also hopefully now know that (cos of even functions) \[\large \int\limits_{-5}^{0}f(x) dx = \frac{ 1 }{ 2 }\left( \int\limits_{-5}^{5}f(x) dx \right)\]

Answer 10

20

Answer 11

But the way I was originally going to show you was more geometric, but then I got confused myself and lost where I was going... but now I remembered again.

Answer 12

np, are you a student btw?

Answer 13

No, i'm a teacher and tutor

Answer 14

Notice if we cut the region from -5 to 5 down to -5 to 0...

it's somewhat obvious the avg value will stay the same, right? Remember that avg value is essentially the average "height of the function" (even if my drawing kinda sucks at showing that... as i just realized)