Find the area of the figure using the limit process for:y = x², y =x, x=0, x= 2.

Answered: Find the area of the figure using the…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter8: Areas Of Polygons And Circles

Section8.4: Cicumference And Area Of A Cicle

Problem 21E: Let N be any point on side BC of the right triangle ABC. Find the upper and lower limits for the...

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Question

Find the area of the figure using the limit process for:
y = x², y =x, x=0, x= 2.

Expert Solution

Step 1

Given,

$y = x^{2}, y = x, x = 0, x = 2$

Step 2

From the figure,

On solving the equation $y = x^{2}, and y = x$ , we get

$x = 1, and y = 1$

So, $y = x^{2}, and y = x$ intersect at the point (1,1)

Then the area of the region bounded by the given curves from $x = 0 to x = 2$ can be written as

Bounded area, A= $A_{1} + A_{2}$

Where , $A_{1} = \int_{0}^{1} (x - x^{2}) d x$ and $A_{2} = \int_{1}^{2} (x^{2} - x) d x$

$The area A_{1} and A_{2} can be evaluated as the limit of a sum$

Step 3

By definition of the limit of a sum

$\int_{a}^{b} f (x) d x = h \lim_{n \to \infty} [f (a) + f (a + h) + . . . . . . . . . f (a + (n - 1) h)]$

Where, $h = \frac{b - a}{n}$

So,

$A_{1} = \int_{0}^{1} (x - x^{2}) d x = \frac{1}{n} \lim_{n \to \infty} [f (0) + f (\frac{1}{n}) + f (\frac{2}{n}) + . . . . . . . . . . + f ((n - 1) \frac{1}{n})] = \frac{1}{n} \lim_{n \to \infty} [0 + (\frac{1}{n} - \frac{1}{n^{2}}) + (\frac{2}{n} - \frac{2^{2}}{n^{2}}) + . . . . . . . . . . + \frac{n - 1}{n} - \frac{{(n - 1)}^{2}}{n^{2}}] = \lim_{n \to \infty} \frac{1}{n} [\frac{1}{n} (1 + 2 + 3 + . . . . . (n - 1)) - \frac{1}{n^{2}} (1^{2} + 2^{2} + 3^{2} + . . . . . . . {(n - 1)}^{2})] = \lim_{n \to \infty} \frac{1}{n} [\frac{1}{n} (\frac{n (n - 1)}{2}) - \frac{1}{n^{2}} (\frac{(n - 1) n (2 n - 1}{6})]$

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