Rahul A.

asked • 05/12/21

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)).


1 Expert Answer


Daniel B. answered • 05/12/21

4.6 (21)

A retired computer professional to teach math, physics

Rahul A.

Yes but 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. When we plot the graph of -cos x(the integral graph) we see that we are just subtracting the y-values i.e (-cos (pi))-(-cos(0)). Am I right?


Daniel B.

My answer is after the dashed line.


Rahul A.

Ok thank you sir!


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