Find the area above x-axis for given function on given interval by using the Fundamental theorem of Calculus.
F(x)= 1+2sinx on [0,pi/2]

- anonymous

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- anonymous

@DanJS

- anonymous

which ftc should I use first or the second?

- DanJS

is this just to integrate with respect to x over that interval?

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## More answers

- DanJS

F(upper)-F(lower)

- anonymous

I don't think so. I believe it has already integrated considering it uses an upper case F(x).. \[F(x)=\int\limits f(x)dx\]Probably just plug in the limits

- DanJS

from 0 to half pi, sin(x) is always a positive number, the function is always positive

- DanJS

do the integral of the function from 0 to half pi

- anonymous

find the area above x-axis- (for given function on given interval)

- DanJS

F(x) is the given function, the area under that is the definite integral value for the interval

- anonymous

I wasn't sure if the notation was specific. I was taught that it was, but I guess it may not be.

- DanJS

yeah that is what integrating will do, it can give you an exact area value for spaces that are curved in any way on the boundries

- anonymous

it has closed brackets towards it so I guess just teach me the way u is right I will check about that with my prof tmr

- DanJS

you see the pictures of 5 boxes, 10 boxes, infinite boxes, to estimate the area under curves

- anonymous

Yeah, I'm aware. But I was afraid that the question already presumes that it was integrated and just wants the student to plug in the limits, hence the "using fundamental theorem of calculus" part.

- anonymous

If notation is specific, then
\[\large \int\limits f(x)dx=F(B)-F(A)\]
I feel like we already have the left hand, now we just need to plug in the limits. But I could be wrong.

- anonymous

you see the pictures of 5 boxes, 10 boxes, infinite boxes, to estimate the area under curves- yes, I know that

- anonymous

but what are u trying to say I am confused now

- DanJS

the given function is F(x) = 1 + 2sin(x), they want the area between x-axis and that function.
That is what the fundamental theorem will do, you need to integrate, Just make the next F even larger. haha

- anonymous

X)

- anonymous

okay by using the first fundmental theorem?

- DanJS

\[\int\limits_{0}^{\pi/2}F(x)dx \]

- anonymous

that's the second i think

- DanJS

oh, you talking about that derivative of an integral , other thing?

- anonymous

no the ftc- I think the π/2F(x)dx u gave above is the second fundamental theo of calc

- anonymous

The second fundamental theorem of calculus is:
\[F(x)=\int\limits_{0}^{x}f(t)dt \implies \frac{ d }{ dx }[F(x)]=f(x)\]

- anonymous

@DanJS okay what is next plz explain me in terms of math I get confused when u talk in sentences.

- anonymous

X)

- DanJS

Integrate the function and you get
x - 2*cos(x)
then you take the upper bound and the lower bound and evaluate those in that x-2cos(x)
take the value of the upper - value of the lower, that is the answer

- DanJS

|dw:1447318077380:dw|

- anonymous

okay i get the upper bound and lower boun dstuff but how did u got
Integrate the function and you get
x - 2*cos(x)?

- DanJS

|dw:1447318108190:dw|

- DanJS

there are properties for the integrals, similar to the way you do limits, i think
so you can re-write addition of terms, as the addition of 2 integrals

- DanJS

it would be the integral of 1 + integral of 2 sinx
then you can move the constant 2 out front the second if you want

- DanJS

remember those limit properties

- anonymous

okay I will look into my notes for that. Plz keep doing the problem after finding the lower and upper bound.

- DanJS

http://www.sosmath.com/calculus/integ/integ02/integ02.html

- DanJS

|dw:1447318458871:dw|

- DanJS

remember cos(pi/2) = 0 and cos(0) = 1
you get an answer of pi/2 + 2

- anonymous

1.571 +2

- DanJS

i would leave it as exact value

- anonymous

or pi/2 +2 yeah that's what i got

- anonymous

okay so the upper and lower bound- u said that it would be (1+2sinx)dx
and the other (x-2cosx) I am confused on the upper/lower bound stuff that u said

- DanJS

oh, just the ends of the interval

- DanJS

you are only considering the function over a restricted bounded region 0 to half pi

- anonymous

oh okayy

- DanJS

the work required is just what is in those draw boxes really, less you need to show more integrating properties, since that is the topic you are learning now

- anonymous

right I understand! Okay thank you. There is another one with the same question but it's askin gfor F(x)= 1-x^2 on [0,2] I will open that into a new question if u wan t

- DanJS

maybe show that the function is continuous and defined for all values on the interval, all the requirements for integrating

- DanJS

nah that's ok

- anonymous

thank you so much!

- DanJS

1-x^2 is a parabola vertex at (0,1) and going down
you will have some area above and some below the x axis,

- DanJS

yeah, maybe show the function is always positive in that interval

- DanJS

state at the end, that the value is the area they want, not sure, depends on who will be reading it

- DanJS

i am falling asleep.. hah it is after 4am here

- DanJS

gonna take off for now, prolly be on tomorrow

- anonymous

how do i show that the fuction is always positive? yeah we will do the next one tmr lets just finish the first one

- anonymous

yeah that's okay we can do that one tmr but like u were mentioning- to show it's positive how do I do that?

- DanJS

sin(x) is + for the whole first quadrant of angles, 0 to pi/2

- anonymous

yeah I know but where do i put that or hwo do i put that in the work I hae so far?

- DanJS

the second one is the same process
integrate 1-x^2 from 0 to 2, the thing is continuous and defined over that interval

- DanJS

[x - (1/3)x^3] from 0 to 2

- anonymous

okay. we can do that one tmr it's okay

- DanJS

k, cya

- anonymous

thanks :) bye tc

- anonymous

the second one is the same process
integrate 1-x^2 from 0 to 2, the thing is continuous and defined over that interval
[x - (1/3)x^3] from 0 to 2
okay, how did u got [x - (1/3)x^3]? @DanJS

- DanJS

You can integrate each term , a property of integrals
\[\large \int\limits_{0}^{2}1*dx - \int\limits_{0}^{2}x^2dx\]

- DanJS

oh yeah i forgot, i remembered i have something, one sec

- anonymous

okay I understnad that now

- anonymous

but from that how did u got [x - (1/3)x^3

- DanJS

summary of lots of things, some arent covered till next semester though, calc 2

##### 1 Attachment

- anonymous

wow let me print that and put in in ma binder :)

- DanJS

integration is reverse of differentation
if you have a function f(x) , you can find the derivative f ' , the integral will move you the other direction , from f ' to the function f

- DanJS

the integral of 1 is the thing that when you take its derivative is 1, kind of like reverse of th epower rule for derivatives, 1 is a constant, or you can say x^0 power
what do you take the derivative of to get x^0 or 1?
x, the integrating rule thing you can remember is one of those handful of common integrals, there are a bunch listed on page 1 of that pdf

- DanJS

integrating a constant is the first one,

- anonymous

it says kx+c if you are taking about the cconstact

- DanJS

right if you integrate some number with respect to a variable, you can slide in that variable to the zero power as multiplied on i guess, x^0 = 1
using the power rule, (second on that list), the integral of 1*x^0 = (1/1) * x^1 or just X

- DanJS

so integrating 1, results in X

- DanJS

or any other constant k, number

- DanJS

just remember to increase power of x by 1, from 0 to 1, the x appears

- DanJS

it will become just like doing derivatives of things like 3x^2, you just remember to multiply by the power and reduce it by 1, the power rule of derivatives..
after a bit, you just do the pattern

- anonymous

okay so 1-x^2 how did u put that in context ans got -----> x - (1/3)x^3???

- DanJS

integrating x^n, you just increase the power by 1 and divide by that same value
integral x^4 ----> a fifth of xto the fifth

- anonymous

wait am reading what u sai d to understna d

- anonymous

so x^2 would be x^3

- DanJS

for the 1 in the prob, you can just remember for constants to add an x when integrating
integral of x^0 ---->x/1 = x

- DanJS

To integrate a power term, it is the second common integrals in the chart
you just increase the power by 1, and also divide by that same value, 1 more than the power

- DanJS

\[\huge \int\limits x^n dx = \frac{ x^{n+1} }{ n+1 }\]

- anonymous

x^3/4 correct?

- anonymous

sorry divided by 3

- DanJS

The first part of the integral in the problem is 1
you can also think of that as the same as x^0

- DanJS

integrating x^0 following that rule, you get x / 1, just an x

- anonymous

right I got it now

- DanJS

dont really need to include the integration constants in definite integrals,

- DanJS

they will fall out at the end anyways

- anonymous

just to put everything togather can u show all the steps mathametically to show how u got
x - (1/3)x^3
u got x from the "1" and how is (1/3)x^3? we got x^3/3 how did it turned into (1/3)x^3?

- DanJS

those are both the same

- anonymous

oh okay! Okay I finally got that part

- DanJS

i just put the n-1 under the variable, or you can make it a fraction and multiply 1/(n+1)

- DanJS

you can see why you have to divide, look at x^5
the derivative is 5x^4
the integral of that should take you back to x^5

- anonymous

yeah i see it now. Okay

- DanJS

you have to divide by that 5, and also increase the power

- DanJS

i started remembering the integrals of the trig functions, by just thinking of it the other way, thinking what was differentiated to get this thing we have to integrate

- DanJS

like the integral of sin(x), is whatever you differentiate to get sin(x),

- anonymous

okay

- anonymous

but hmm just explain me everything in math/ the equation u had with "n" that's much clear instead of started remembering the integrals of the trig functions, by just thinking of it the other way, thinking what was differentiated to get this thing we have to integrate

- anonymous

so after that we get x - (1/3)x^3 right! @DanJS

- DanJS

ok, yes that is right, you can include the integrating constants if you want, they will cancel at the end when you evaluate the values

- anonymous

and my answers is x-(1/3)x^3

- DanJS

yes, it is from 0 to 2, to get the areas

- anonymous

so there were basically 3 steps for thsi answers and I will include what we discussed about getting x-(1/3)x^3 asmy answer in my notes

- DanJS

|dw:1447373372714:dw|

- DanJS

|dw:1447373499951:dw|

- DanJS

-2/3

- DanJS

The definite integral gives you the NET area between f(x) and the x-axis... with the area above axis positive, and area below axis negative

- anonymous

okay so after finding x-(1/3)x^3 u plug in 2 and 0 as x and find the final answer -2/3. okay that's what e did in the previous one

- DanJS

yeah the fundamental theorem thing
\[\int\limits_{a}^{b}f(x)dx = F(b) - F(a)\]

- anonymous

yes i get it now tysm for taking ur time. but tbh I really do not understand when someoen explains math in sentences to me I only learn when it is explained directly. but tysm for explaining makes better sens enow tysm

- DanJS

i have to take off for a hour or so in a minute

- DanJS

ok, i will do that, i just hate typing in the commands.. i will from now on... just was faster typing words.. lol sorry bout that

- anonymous

yeah we are done with this question so u are good to go for now lol

- DanJS

:) , k , be back later

- anonymous

bye thanks

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