- #1

- 52

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[itex]\int^{a}_{b}[/itex] s(x) dx = [itex]\sum^{n}_{k=1}[/itex] s[itex]_{k}[/itex] [itex]\times[/itex] (x[itex]_{k}[/itex] - x[itex]_{k-1}[/itex])

and says that to compute the integral we multiply each constant value s[itex]_{k}[/itex] by the length of the k'th subinterval and add together all these products.

I have understood till this point but he further says that the values of s at the subdivision points are immaterial since they do not appear on the right hand side of the equation.

i.e. if s is constant on the open interval (a,b) say s(x) = c if a<x<b then

[itex]\int^{a}_{b}[/itex] s(x) dx = c[itex]\sum^{n}_{k=1}[/itex] (x[itex]_{k}[/itex] - x[itex]_{k-1}[/itex]) = c(b-a) regardless of values s(a) and s(b). If c>0 and s(x) =c for all x in the closed interval [a,b], the ordinate set of s is a rectangle of base b-a and altitude c; the integral of s is c(b-a) the area of the rectangle.

I am not able to understand the statement "changing the value of s at one or both endpoints a or b changes the ordinate set but does not alter the integral of s or the area of its ordinate set."

If there is change in the ordinate set shouldn't there be a change in the area too? Also if the end points are ignored wouldn't the area change? Is it because we are computing the area of an open interval?