$$\int\limits_{-2}^{k} (4-2x) dx = 15 $$

where k1 find the value of k

Question

$$\int\limits_{-2}^{k} (4-2x) dx = 15 $$    

where k>1 find the value of k

jtvatsim · Answer

This one isn't too bad. You just need to take the integral like usual and everything will end up working itself out in the end. :)

jtvatsim · Answer

Do you know the integral of (4 - 2x) ?

anonymous · Answer

[(4x/x - 2x^2/2)] is it like this?

jtvatsim · Answer

you are really close. The second part ... - 2x^2/2 is correct but the first part 4x/x is not right.

anonymous · Answer

4x only

jtvatsim · Answer

Aha! Indeed, that is correct!

jtvatsim · Answer

Alright, so what we have right now is this $$[4x - x^2]_{-2}^k = 15$$ does that make sense to you? :)

anonymous · Answer

u took out 2 from this " - 2x^2/2 "

anonymous · Answer

yes, it do make sense.

jtvatsim · Answer

yes, because -2x^2 divided by 2 is just - x^2. Good.

jtvatsim · Answer

OK, so now expand the brackets or whatever you want to call it and get: $$(4k - k^2) - (4[-2] - [-2]^2) = 15$$ okay so far?

jtvatsim · Answer

This is just what you always do when evaluating integrals. :)

anonymous · Answer

okay. let me try solve it fist.

anonymous · Answer

*first

jtvatsim · Answer

sounds good, let me know what you get. :)

anonymous · Answer

ok

anonymous · Answer

I get k = 1 and k=3. but it says, k>1 so it means, k= 3 is the answer. is my theory right?

jtvatsim · Answer

Yahoo! You did it! Congrats! :D

anonymous · Answer

TQ, it is your effort too, to teach me the step. so congratz to u too.. :D

jtvatsim · Answer

You are welcome. :)