Question

Express as a single integral (can someone explain this to me?)

Answer 1

*tick tock tick tock*
lol

Answer 2

Look at this page and see if it's helpful at all...

http://www.wolframalpha.com/input/?i=single+integral

Answer 3

lol Rogue:P

Answer 4

I'm posting it one second

Answer 5

The middle one is just 0 since the limits of integration are the same.

Answer 6

Well, that's weird...

Answer 7

do they just adopt the largest interval possible? Do you know where i can find a rule set for these problems?

Answer 8

Properties of a definite integral
http://www.sosmath.com/calculus/integ/integ02/integ02.html

Answer 9

is there something weird about this question but yeah man thanks you are saving my retrice I have a general intermediary metabolism test in a couple days that I haven't started studying for so I need to learn this stuff fast

Answer 10

Ugh, bio, so much memorization :(

I have an idea on how to do this one, give me a few.

Answer 11

Okay, I got it!

Answer 12

I dont care about the answer more about the journey to the answer.

Answer 13

man that was wrong!!

Answer 14

\[\int\limits_{0}^{2} f(x) dx - \int\limits_{4}^{4} (f(x) + g(x))dx - \int\limits_{2}^{4} f(x) dx\]\[\int\limits\limits_{0}^{2} f(x) dx - \int\limits\limits_{2}^{4} f(x) dx\]\[\int\limits\limits\limits_{2}^{4} f(x-2) dx - \int\limits\limits\limits_{2}^{4} f(x) dx = \int\limits_{2}^{4} (f(x-2) - f(x))dx\]

Answer 15

I don;t follow rogue

Answer 16

:( Okay, where do you get lost?

Answer 17

from the start lol

Answer 18

Okay, the start is your problem, lol.

Answer 19

oh I see it becomes 0

Answer 20

because it is a =4 b =4

Answer 21

yeah, same limits of integration = area of a line = 0

Answer 22

I'm lost when it comes to subtracting integrals though

Answer 23

does it just become the largest interval possible?

Answer 24

You can combine integrals if they have the same limits. Ex.\[\int\limits_{a}^{b} g(x)dx + \int\limits_{a}^{b} h(x)dx = \int\limits_{a}^{b} \left[ g(x) + h(x) \right]dx\]

Answer 25

right I understand that rule :)

Answer 26

Okay, as per your 'largest interval possible,' yes, if its something like this\[\int\limits_{a}^{b} h(x)dx + \int\limits_{b}^{c} h(x)dx = \int\limits_{a}^{c} h(x)dx\]

Answer 27

But we don't have that, and we have different limits of integration. So in order to make combine it and make it 1 integral, we need same limits of integration.

Answer 28

To get the limits from 0 to 2 of \[\int\limits\limits\limits_{0}^{2} f(x) dx\]to 2 to 4, we translate the function 2 to the right.

Answer 29

oh I see, so I need to learn the basic function transformation rules from a long time ago :)

Answer 30

If you have a graphing calculator, play around w/ y = x and y = x - 2. The area under y = x from 0 to 2 is the same as the area under y = x - 2 from 2 to 4.

Answer 31

Its the same area, just with different limits of integration, which we can be combined w/ the other integral because then they have the same limits.

Answer 32

oh I see, we are just shifting the function, so with negative integrals we need to transform the function so that the intervals are the same so we can combine them?

Answer 33

Is it the same with positive integrals

Answer 34

No, disregard the negative/positive. In order to combine 2 integrals, they need to have the same limits of integration, that's the main point.\[\int\limits_{a}^{b} f(x)dx + \int\limits_{a}^{b} g(x) dx = \int\limits_{a}^{b}\left[ f(x) + g(x) \right] dx\]\[\int\limits\limits_{a}^{b} f(x)dx - \int\limits\limits_{a}^{b} g(x) dx = \int\limits\limits_{a}^{b}\left[ f(x) - g(x) \right] dx\]

Answer 35

ok I think I get it now, just to clarify though

you can add integrals that are within the limits of another the other integral so in the case you pointed out.

c > b
and
a < b

Answer 36

or am I just lost