graph of integral function.
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)).