Question

Maths question definite integartion sum

Answer 1

Answer:

Step-by-step explanation:

To understand this, let’s evaluate the area PRSQP between the curve y = f(x), x-axis and the coordinates ‘x = a’ and ‘x = b’. Now, divide the interval [a, b] into ‘n’ equal sub-intervals denoted as:

[x0, x1], [x1, x2], [x2, x3] …. [xn – 1, xn], where,

x0 = a, x1 = a + h, x2 = a + 2h, x3 = a + 3h ….. xr = a + rh and xn = b = a + nh

Or, n = (b – a)/h. Note that as n → ∞, h → 0.

Now, the region PRSQP under consideration is the sum of all the ‘n’ sub-regions, where each sub-region is defined on subintervals [xr – 1, xr], r = 1, 2, 3 … n. Now, look at the region ABDM in the figure above. We can make the following observation:

Area of the rectangle (ABLC) < Area of the region (ABDCA) < Area of the rectangle (ABDM) (1)

Also, note that as, h → 0 or xr – xr – 1 → 0, all these three areas become nearly equal to each other. Hence, we have

sn = h [f(x0) + f(x1) + f(x2) + …. f(xn – 1)] = h r=0∑n–1 f(xr) … (2)

and, Sn = h [f(x1) + f(x2) + f(x3) + …. f(xn)] = h r=1∑n f(xr) … (3)