   Chapter 4.2, Problem 62E

Chapter
Section
Textbook Problem

Finding Area by the Limit Definition In Exercises 57-62, use the limit process to find the area of the region bounded by the graph of the function and the y-axis over the given y-interval. Sketch the region. h ( y ) = y 3 + 1 , 1 ≤ y ≤ 2

To determine

To calculate: The area of the region bounded by the function f(y)=y3+1, in the interval [1,2] and the y -axis.

Explanation

Given: The provided function is,

f(y)=y3+1, in the interval [1,2].

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

i=1ni2=n(n+1)(2n+1)6

The sum of first n natural numbers is given by the formula,

i=1ni=n(n+1)2

Sum of a constant n times is written as,

i=1nc=nc

The sum of cubes of first n natural numbers is given by the formula,

i=1ni3=n2(n+1)24

Using upper endpoints area is,

A=limni=1nf(Mi)(Δx)

Here, Mi is the upper endpoints.

Calculation: The provided function is,

f(y)=y3+1

We will differentiate f(y) with respect to y and obtain f'(y),

f'(y)=3y2

Now, equate f'(y) to zero to get,

3y2=0y=0

So, the slope of the function w.r.t y -axis is zero at one point. Check where the curve is approach by put y= and y=.

On put y=

f()=3+1

So, f() is approach to .

Now put y=

f()=3+1

That means the curve is approaching to as y.

Therefore, the required graph with the region is shown below,

As the given function f(y) is continuous and non-negative in the interval [1,2]. So, by the partition of interval into n subintervals each of equal width.

Therefore, considering the width of each rectangle to be (Δy) then,

The width of each rectangle is,

Δy=21n=1n

The area can be calculated by the use of lower endpoints (mi) or upper endpoints (Mi). For this problem the upper endpoints are convenient.

Now, find upper endpoints (Mi).

Mi=1+in=i+nn

So, the upper endpoint is Mi=i+nn.

Therefore, apply the formula to find the area (A) of the region,

A=limni=1nf(Mi)(Δy)

Now, substitute the values to find the area of the region

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