Chapter 5.2, Problem 51E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Finding Area by the Limit Definition In Exercises 47-56, use the limit process to find the area of the region bounded by the graph of the function and the x-axis over the given interval. Sketch the region. y = 25 − x 2 ,   [ 1 ,   4 ]

To determine

To calculate: The area of the region bounded by the function y=25x2, in the interval [1,4] and the x-axis.

Explanation

Given: The provided function is:y=25âˆ’x2, in the interval [1,4].

Formula used: The sum of squares of first n natural is given by the formula:

âˆ‘i=1ni2=n(n+1)(2n+1)6

The sum of a constant n times is written as: âˆ‘i=1nc=nc

The sum of first n natural numbers is given by the formula âˆ‘i=1ni=n(n+1)2

Use right endpoints area is written as: Area=limnâ†’âˆžâˆ‘i=1ny(Mi)(Î”x),

Calculation: Note that the provided function y is continuous and non-negative in the interval [1,4].

So, begin by partition the interval into n subintervals each of width Î”x:

Î”x=4âˆ’1n=3n

Area can be calculated by left endpoints (mi) or right endpoints (Mi). For this problem the right end points are convenient.

Right endpoints (Mi) are:

1+3in=3i+nn,

i=â€‰1,2,3,.......,n

So, Area=limnâ†’âˆžâˆ‘i=1ny(Mi)(Î”x)

Where, Mi are the right endpoints, and Mi=3i+nn.

So, Area=limnâ†’âˆžâˆ‘i=1ny(3i+nn)(Î”x)

Now, put value of y(3i+nn),

Area=limnâ†’âˆžâˆ‘i=1ny(3i+nn)(Î”x)=limnâ†’âˆžâˆ‘i=1n(25âˆ’(3i+nn)2)(3n)

Split the expression in parts to use summation formulas:

limnâ†’âˆžâˆ‘i=1n(25âˆ’(3i+nn)2)(3n)=limnâ†’âˆžâˆ‘i=1n75nâˆ’limnâ†’âˆžâˆ‘i=1n3n3(3i+n)2

Expand the second sum: limnâ†’âˆžâˆ‘i=1n75nâˆ’limnâ†’âˆž

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