Question

Let f(x) = Integral cos xdx for 0 12 years ago

Answer 1

so what is the integration?

Answer 2

cos xdx?

Answer 3

thats the arguement. f(x) equals the integral of cos x dx. What does the integration give us?

Answer 4

Total area under the curve?

Answer 5

still trying to work out the rest of what you are asking really; can you provide a better representation of the question? ascii doesnt do well with it

Answer 6

sin(x) + C derives to cos(x) so we have an antiderivative to play with; but the information in the question is not making a lot of sense to me at the moment wihtout some clarification

Answer 7

\[Let f(x) = ∫\cos x dx for 0 < x < \pi. If f ( \pi/6) = 1, then f ( \pi / 2) ?\]

Answer 8

That's exactly what the question stated.

Answer 9

if you can take a screenshot or picture of the question and attach it to your post, it would be easy to see if you are presenting the information correctly to us.

what im used to seeing is something like: \[f(x)=\int_{0}^{h(x)}g(t)~dt\]
or some such ....

Answer 10

The question that I stated above. is correct. There isn't any information given other the that..

Answer 11

I think I will have take 2 integrals. One from pi/6 to pi and then another one from pi/2 to pi

Answer 12

otherwise, all im reading this as is:
\[f(x)=\int_{0}^{\pi}~cos(x)~dx=0\]
therefore f(x) = 0 for any x between 0 and pi ....

Answer 13

\[\int\limits_{\pi/6}^{\pi} - \int\limits_{\pi}^{\pi/2} \]

Answer 14

And then then take the sin of those variables and then add them together? Would it work?

Answer 15

if its an intial value problem such that the antiderivative of cos(x) = sin(x)+C; then sin(pi/6)+C=1

sin(pi/2) - sin(pi/6) = ?

youre new rendition is even stranger ... which is why im thinking this might be simpler if you could provide a "hardcopy" of the question to make sure.

Answer 16

http://www.mrefram.com/apcalc/APCalcReleasedFinal.pdf. Gp that website and look at number 8.

Answer 17

the you will find the question

Answer 18

its saying its a bad link ....

Answer 19

http://www.mrefram.com/apcalc/APCalcAssignments.htm

Answer 20

Then click on practice final review. If possible..

Answer 21

got it

Answer 22

Alright :D

Answer 23

is that they answer key below it?

Answer 24

Yes, the answer is suppose to be 3/2

Answer 25

what i do notice is that it is an indefinite integral; no limits. So im thinking my initial value idea is more spot on.

Answer 26

ok.

Answer 27

\[f(x)=\int~cos(x)~dx\]
\[f(x)=sin(x)+C\]
\[when~x=pi/6:~sin(pi/6)+C=1\]
this allows us to find a suitable value for C; C = 1 - sin(pi/6) therefore:\[f(x)=sin(x)+1-sin(pi/6)\]

Answer 28

when x=pi/2, what do we get?

Answer 29

2

Answer 30

I got 3/2 for the answer

Answer 31

me too; the key was in understanding what type of integral they were asking for.

a definite integral with limits from a to b defines the area underneath a curve.

an indefinite integral provides us a family of antiderivatives which we can use an initial value on to find a specific function to play with.

Answer 32

Alright. Thank you so much. I got it :D

Answer 33

youre welcome :)