graph of integral function.

Question

Pls Draw Sin x and -Cos x Graphs Before Reading.(to understand my query,sorry)

we know the sin x graph gives the slope of -cos x at every point.

when we calculate the average of -cos x graph i.e (-cos (pi))-(-cos(0))/pi-0 we calculate the average for the graph -cos x. join the points on graph of -cos x. we get 2/pi.

so in short what we are doing here is calculating average slope.

now this is similar to calculating and adding up all the values on the graph of sin x from 0 to pi and dividing it by number of samples

i.e [INTEGRAL FROM 0 TO PI] sin x DIVIDED BY pi/dx

where no. of samples is pi/dx = length of interval/ spacing between points.

rearranging we get [INTEGRAL FROM 0 TO PI] Sin x dx/pi, which when integrated we get 2/pi

so its like [INTEGRAL FROM 0 TO PI] Sin x dx/pi = (-cos (pi))-(-cos(0))/pi-0 .

both LHS and RHS signify and convey that we have calculated average

similarly when we calculate normal integral here

say [INTEGRAL FROM 0 TO PI] Sin x dx = (-cos (pi))-(-cos(0))

we get the answer as 2.

SEE integral of sin x dx means we add all the thin rectangles sin x dx under sinx graph from 0 to pi and we get 2

and as we see in the figure in the sin x graph this is area under the curve.

So what does the RHS i.e (-cos (pi))-(-cos(0)) which we get as 2, convey/ express? how does it express that we have calculated area or what does that part tell us from the graph of -cos x. from the graph of -cos x it just seems we made subtraction of the y values of (-cos (pi))-(-cos(0)).

THANK YOU.

Answer 1

You wrote

"the sin x graph gives the slope of -cos x at every point".

You could have also chosen to say

"the sin x graph gives the slope of 1-cos x at every point".

Or you could have chosen any other constant besides 1.

But I think the choice of 1 illustrates your point the best.

Notice that 1-cos(π) = 2, which is the area under the sin curve between 0 and π.

The function F(x) = 1-cos(x) is most illustrative because F(0) = 0.

That mean, that F(x) represents the accumulated area as you keep increasing

your range starting from 0.

In contrast, the function 100-cos(x) would represent the accumulated area + 99.

And the function -cos(x) represents the accumulated area - 1.

------------------------------------------------

Yes, you are right, we are just subtracting the y-values.

Let me try to explain by answering three questions.

Only Question #2 is the one you asked.

Question #1:

"What does it say on the graph of 100-cos x, the RHS i.e (100-cos (pi))-(100-cos(0)).

On 100-cos x graph how does it express that we have calculated area?"

Answer:

The function 100-cos(x) records how we accumulate area under sin(x) when going from 0 to π.

But at the beginning, when x = 0, we have been given a head start of 99.

So by the time we get to π, 100-cos(π) records not only the gain between 0 and π, but also the

head start of 99.

The difference (100-cos (π))-(100-cos(0)) gives us the net gain between 0 and π,

conveniently cancelling out the initial head-start.

Question #2:

"What does it say on the graph of -cos x, the RHS i.e (-cos (pi))-(-cos(0)).

On -cos x graph how does it express that we have calculated area?"

Answer:

The function -cos(x) records how we accumulate area under sin(x) when going from 0 to π.

But at the beginning, when x = 0, we have been given a handicap of -1.

So by the time we get to π, -cos(π) records not only the gain between 0 and π, but also the

handicap of -1.

The difference (-cos (π))-(-cos(0)) gives us the net gain between 0 and π,

conveniently cancelling out the initial handicap.

Question #3:

"So in what way is -cos(x) special."

Answer:

The function -cos(x) is special only in that it has the integration constant of 0.

The function 1-cos(x) is special in a different way: It actual records the area under sin(x)

without any head start or handicap.

The reason is that 1-cos(0) = 0.

Having the initial value of 0 makes the function special in that the area under the curve is simply

1-cos(π).

The function -cos(x) would be equally special if we wanted the area under sin(x) starting from π/2.

Does this help?