Question

# Which of the following gives the area of y = (/x^{2} /) between x = 0, x = b and x-axis? where k varies from 1 to n and n tends to infinity where k varies from 1 to n and n tends to infinity\sum \frac{b^{3}}{n^{3}} k^{3}where k varies from 1 to n and n tends to infinity where k varies from 1 to n and n tends to infinity

Solution

## The correct option is B  where k varies from 1 to n and n tends to infinity We want to find the area of the parabola between x = 0 and x = b  . All the options are given as sum of infinite terms. As we learned in Archimedes’s method for finding the area of a curve, we will divide the given curve into n intervals of width /( \frac{b-0}{n} /). In each interval we will construct a rectangle with width as the length of interval and height as the value of the function on the rightmost point of the interval(when we approach from top).  We will now calculate the area of the rectangle formed. For example, if we consider the interval [\frac{b}{n}, ~ \frac{2b}{n}], corresponding to the second rectangle, the height of the rectangle will be f(2bn), where f(x) = x2. Similarly the height of kth rectangle will be f\frac{kb}{n}, Area of this rectangle will be bnf(kbn,)=bn(kbn)2 =b3n3k2 When n tends to infinity, the error or extra area reduces to zero. So we can say the sum of this expression, where k varies from 1 to n, when n tends to infinity will give the area of the curve.  So the answer will be ∑b3n3k2, where k varies from 1 to n and n tends to infinity.

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