/Intro2IntegralCalculus(Area)

INTEGRAL CALCULUS: AN INTRODUCTION Integral

Calculus is, as we know, the second and the not-so-straightforward branch of Calculus. However, in this

section, we will attempt to simplify some issues by introducing the two major factors on which integral calculus is based: AREA and DISTANCE. We begin with area.

THE CONCEPT OF 'AREA' How do we define area? The answer depends on the figure we're dealing with. For solid shapes, we have explicit formulas for determining their areas; for a triangle, the area is half the base times height. The area of a rectangle is the length times the width. The area of a parallelogram is the base times the height, and so on. But how do we find the area of a region with curved sides? Well, the process of finding such area is not very straightforward. So now, the problem is to obtain a precise definition of area, and we shall start by performing a procedure which is similar to that used in defining a tangent: approximating slopes of secant lines and then taking the limit of those approximated slopes. In the same vein, we shall approximate the area of a given region P by using a given number of rectangles (whose individual areas will be calculated). Then we take the limit of the areas of the rectangles as the number of rectangles increase. This procedure is generalized below.

GENERAL PROCEDURE FOR COMPUTING AREAS Suppose we want to find the area under the curve y = f(x) from a to b (this area, which is the blue region is represented by P:

We then “partition� P into any number of equal-sized strips (which will form rectangles) by drawing vertical lines x = 1/n, x = 2/n, x = 3/n, x = 4/n................x = n/n (where n is the number of strips that the region P has been divided into. In this case, we'll divide P into 10 equal strips). Note that the rectangles MUST be of equal width. The figure below shows the graph partitioned into 5 subintervals, the curve touches the right hand endpoints of the approximating rectangles:

We move on to approximate the area of each rectangle (using the width and height of each rectangle). The width is equal to the base of the rectangles, that is.

x1 = 1/n, x2 = 2/n, x3 = 3/n, x4 = 4/n Therefore,

xn = n/n, The height of each rectangle is equal to the value of the function y = f(x) at the right endpoints of the subintervals, that is, f(1/n), f(2/n), f(3/n)..............f(n/n). So, each rectangle would have width 1/n and the heights will be f(1/n), f(2/ n), f(3/n)..............f(n/n).

Therefore, the sum A1 of the areas of the approximating rectangles is:

A1 = 1/n.f(1/n) + 1/n.f(2/n) + 1/n.f(3/n) + ....... + 1/n.f(n/n) Equation 1

But, we have a small problem: from the graph, you can see that the area of P is much less than A1, that is,

Area of P < A1 (this is because the rectangles extend way up beyond the region P).

At least, A1 gives a rough estimate of the area of the region P. However, we can do better. Notice that the curve in the graph above touches the right endpoints of the rectangles. We can make a better approximation of the area of P by using rectangles that are considerably smaller than those above; therefore, the curve would now touch the lefthand endpoints:

So, the sum A2 of the areas of the approximating rectangles will be

A2 = 1/n.f(1/n) + 1/n.f(2/n) + .......... + 1/n.f(1â&#x20AC;&#x201C; 1/n) Equation 2a Depending on the function, there may be cases where the first rectangle does not exist, i.e, f(1/n) = 0. In this case, A 2 becomes

A2 = 1/n.0 + 1/n.f(1/n) + 1/n.f(2/n) + ........ + 1/n.f(1â&#x20AC;&#x201C; 1/n) Equation 2b Either way, from the graph, we find that the area of P is now larger than A2. But now, we are very close: we have lower and upper estimates, and thus we can conveniently say that the area of P lies somewhere between A1 and A2:

A1 < Area of P < A2

This procedure can be repeated with an even larger number of strips, thereby increasing our chances of obtaining better estimates.

At this point, you’ll notice that the sums of areas of the rectangles (that is, A 1 and A2, for several values of n) seem to be approaching a particular value. That value will be the limit of the sums of areas of the rectangles, which we will now compute.

Rearranging equation 1 gives

[

1/n f(1/n) + f(2/n) + f(3/n) + ................ + f(n/n)

]

Equation 3 (Don’t forget that the expressions f(1/n), f(2/n), f(3/n), f(n/n) are REAL NUMBERS). Observe that the sum in the squared bracket follows a pattern. This sum can be written compactly using sigma notation:

n

A

=

1/n

∑

f([1+i]/n)

i=0 Following the procedure above, we can therefore say that the area A of a region P is the limit of the sums (upper and lower) of the approximating rectangles.

lim A1

n→∞

and

lim A2

n→∞

(BOTH LIMITS SHOULD YIELD THE SAME RESULT, IF THEY EXIST).

(The upper and lower approximating sums are represented by A1 and A2 respectively). The above limits can be read as

“The limit of the sum of the areas of the approximating rectangles as the number of rectangles increase infinitely is......”

We can modify this procedures to work in finding larger areas: Like the general procedure, we divide the region P into any number of strips. Since we are looking for the area of P between a and b, then the width of the closed interval [a, b] is b – a. So, if there are n strips, then the width of each strip would be

∆x where

=

(b – a)/n

∆x is the width of each strip. The right-hand endpoints of the subintervals are x1

=

a + ∆x,

x2

=

a + 2∆x,

x3

=

a + 3∆x,

x4

=

a + 4∆x

and so on. Therefore, the area of each rectangle will be

An = f(xn)∆x

Where n is ANY REAL POSITIVE NUMBER, and f(x n) is the height of the rectangles; which is the value of f at the right endpoints of the subintervals. So, for instance, the area of the 10th strip would be

A10 = f(x10)∆x e.t.c.

If we use the right-hand endpoints, then the area of P is given by the sum of the areas of these rectangles:

Rn = f(x1)∆x + f(x2)∆x + f(x3)∆x + f(x4)∆x + ...... + f(xn)∆x Equation 4

On the other hand, if we use left-hand endpoints, then the area of P is given by:

Ln

=

f(x0)∆x + f(x1)∆x + f(x2)∆x + ....... + f(xn – 1)∆x Equation 5

The next step is to find the limits of the two sums (equations 4 and 5):

lim Rn

and

n→∞

lim Ln

n→∞

(AGAIN BOTH LIMITS SHOULD YIELD THE SAME RESULT, IF THEY EXIST AND IF f IS CONTINUOUS).

Note that these limits correspond to

lim A1

and

lim Rn

and

n→∞

lim A2

n→∞

also

n→∞

lim Ln

n→∞

represent the upper and lower limits of the sums respectively). Note that these limits exist if and only if f is

continuous. As n increases, we find that the estimates get even better. So, based on the procedure described above, we can thus define area like this:

“If a region P lies under a continuous function f, then the area of such region is the limit of the sum of the areas of the approximating rectangles.”

The definition above can be interpreted as:

A

lim Rn = n→∞

=

lim [ f(x1)∆x + f(x2)∆x + ..... + f(xn)∆x]

n→∞

Equation 6

AND

A

lim Ln = n→∞

lim [ f(x0)∆x + f(x1)∆x + ..... + f(xn – 1)∆x]

=

n→∞

Equation 7

In the procedure described above, the heights of the rectangles are the values of the function f specifically at the lefthand and right-hand endpoints. Alternatively, we can take the heights of the rectangles to be the values of f at ANY POINT within the subinterval (instead of using endpoints). Such points are called sample points and are generally represented as

x* i

Therefore, using sample points, we can define the area A of a region P by

A

=

lim

n→∞

[f(x *)∆x 1

+ f(x2*)∆x + ..... + f(xn*)∆x

]

Equation 8 Notice that equations 6, 7 and 8 are all sums with many terms. Thus, they could be respectively written using sigma notation as follows:

A

A

A

=

=

=

lim

n→∞

lim

n

∑ f(xi)∆ x

i=0

n

∑ f(xi–1)∆ x

n→∞ i=0

lim

n

∑ f(xi*)∆ x

n→∞ i=0

Next, we move on to another fundamental concept of an integral: distance. But before we do, let's study some examples. They illustrate the procedure for finding areas described above.

Note: These examples are actually exercises taken from my textbook.

EXAMPLE 1 (a)

Estimate the area under the graph of f(x) = 1/x from x = 1 to x = 5 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate?

(b)

Repeat part (a) using left endpoints.

Solution (a).

The function in question is the reciprocal function y = f(x) = 1/x. Our task is to find the area under the graph of y = 1/x from x = 1 to x =5 (let this region be represented by M).

The graph above has been divided into 4 strips: S1, S2, S3 and S4. The graph also touches the right endpoints of the strips. From the graph we find that a = x0 = 1

and

b = xn = 5

The question is asking us to estimate the area of the region in the interval [1, 5] using 4 approximating rectangles and right endpoints; this means that the number of strips n = 4. Thus, the width (∆x) of each strip is given by ∆x

=

=

(5 – 1)/4

(b – a)/n

Therefore,the width of each strip is ∆x

=

1

The width of each strip is 1. Next, we need the heights of all four rectangles, which are given by:

x1

=

a + ∆x

=

x2

=

a + 2∆x

=

x3

=

a + 3∆x

=

x4

=

a + 4∆x

=

1+1 1 + 2(1) 1 + 3(1) 1 + 4(1)

=

2

=

3

=

4

=

5

Thus, the area A1 of required region is given by the sum of the areas of the approximating rectangles:

4

A1

∑ f(x )∆ x

lim

=

i

n→∞

i=1

which means

4

A1

lim

=

n→∞

∑ (1/x )∆ x i

i=1

So,

A1

=

f(x1)∆x + f(x2)∆x + f(x3)∆x + f(x4)∆x

A1

=

f(1/x1)∆x + f(1/x2)∆x + f(1/x3)∆x + f(1/x4)∆x

A1

=

(1/2).1 + (1/3).1 + (1/4).1 + (1/5).1

A1

=

(1/2) + (1/3) + (1/4) + (1/5)

A1

=

77/60

»

1.283

Therefore, the area of the region M is approximately 1.283 (using right endpoints). From the graph you'll observe that the sum of the areas of the rectangles is more than the actual area of M ; therefore, A1 is an underestimate.

(b) If we decide to use left-hand endpoints, then the area A2 of M would be:

4

A2

=

∑ f(x

lim

n→∞

)∆ x

i-1

i=1

which means

4

A2

=

lim

n→∞

∑

(1/xi-1)∆ x

i=1

The graph below has been divided into 4 strips: S1, S2, S3 and S4 like the one above. The only difference is that it touches the left endpoints of the strips. Therefore, we have

A2

=

f(x0)∆x + f(x1)∆x + f(x2)∆x + f(x3)∆x

A2

=

f(1/x0)∆x + f(1/x1)∆x + f(1/x2)∆x + f(1/x3)∆x

A2

=

(1/1).1 + (1/2).1 + (1/3).1 + (1/4).1

A2

=

1 + (1/2) + (1/3) + (1/4)

A2

=

25/12 »

2.083

Thus, the area A2 of M is 2.083 (using left endpoints).

We find that the area of M is much less than A2; therefore, A2 is an overestimate. At least, we know that

A1 <

Area of M

<

Area of M

<

A2

In other words, 1.283

<

2.083

Using a graphing calculator, or a scientific calculator, or even a computer, we find that the exact area of M is roughly 1.60944.

EXAMPLE 2 (a) (i) Estimate the area under the graph of f(x) = x3 + 2 from x = -1 and x = 2 using three rectangles and right endpoints. (ii) Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles (b)

Repeat part (a) using left endpoints.

(c)

Repeat part (a) using midpoints.

(d)

From your sketches in parts (a), (b) and (c), which appears to be the best estimate?

Solution (a). (i) Understand the question above. Put simply, it's asking us to estimate the area using left endpoints, right endpoints and midpoints (with 3 and 6 approximating rectangles in each case).

The function here is a polynomial f(x) = x3 + 2 We are to estimate the area A from x = -1 and x = 2 (let this region be T), using three rectangles and right endpoints.

Here is its graph:

So now, we partition T into 3 strips, and let the curve touch each right endpoint of each strip. Again, view the graph on the following page; it shows the graph of f touching the right endpoints of three approximating rectangles.

From the graph, we find that a = -1, b = 2 and the number of strips n = 3 (since there are three rectangles). The width (∆x) of each strip would therefore be

∆x

= (b – a)/n

∆x

= (2 – [-1])/3

∆x

=

3

Therefore, the width of each strip is 1. Also, we’ll need the heights of the three rectangles, which are given by:

x1

=

a + ∆x

=

-1 + 1

x2

=

a + 2∆x

=

-1 + 2(1) = 1

x3

=

a + 3∆x

=

-1 + 3(1) = 2

= 0

The area of T is therefore given by the sum of the areas of the approximating rectangles:

3

A1

=

lim n→∞

which gives

∑ f(x )∆ x i

i=1

3

A1 A1 A1

=

lim

n→∞

∑[(x )

3

i

+ 2]∆ x

i=1

=

1(03 + 2) + 1(13 + 2) + 1(23 + 2)

=

2 + 3 + 10

=

15

Therefore, the area A1 of T is 15 (using right endpoints and 3 approximating rectangles). The question then asks us to improve our estimate by using 6 rectangles. Therefore, we would have to perform the same calculation we did above; only this time, we use the right endpoints of six approximating rectangles:

(a). (ii) The graph below shows the graph of f touching the right endpoints of six approximating rectangles:

From the graph, a = -1, b = 2 and the number of strips n = 6 (since there are now six rectangles). The width (∆x) of each strip is given by:

∆x

= (b – a)/n

∆x

= (2 – [–1])/6

= ½

Therefore, the width of each strip is ½. Now, we’ll need the heights of the six approximating rectangles, which are given by:

x1

=

a+

∆x

=

-1 + ½

=

-½

x2

=

a + 2∆x

=

-1 + 2(½)

=

0

x3

=

a + 3∆x

=

-1 + 3(½)

=

½

x4

=

a + 4∆x

=

-1 + 4(½)

=

1

x5

=

a + 5∆x

=

=

3/2

x6

=

a + 6∆x

=

=

2

-1 + 5(½) -1 + 6(½)

So, therefore, the area A2 of T is given by the sum of the areas of the approximating rectangles:

6

A2

∑

lim

=

n→∞

f(xi)∆ x

i=1

6

A2

lim

=

n→∞

which gives

∑ [(x )

3

i

+ 2]∆ x

i=1

A2

=

½ [(-½)3 + 2] + ½ [(0)3 + 2] + ½ [(½)3 + 2] + ½ [(1)3 + 2] + ½ [(3/2)3 + 2] + ½ [(2)3 + 2]

A2

=

½ [(-1/8) + 2] + ½ [0 + 2] + ½ [(1/8) + 2] + ½ [1 + 2] + ½ [(27/8) + 2] + ½ [8 + 2]

A2

=

½[15/8] + ½[2] + ½[17/8]

A2

=

15/16 + 1 + 17/16 + 3/2

A2

=

195/16

=

+ +

½[3]

+ ½[43/8]

+

½[10]

43/16 + 5

12.1875

Using 6 rectangles, the area A2 of T has been narrowed down to 12.1875.

Summary of this first part: When we used three rectangles and right hand endpoints, the area of the region T (x = -1 and x = 2) is 15, and, when we used six rectangles and right hand endpoints, the area is now 12.1875.

Next, we will repeat parts (a.1) and (a.2) above, but this time, we'll be using LEFT HAND ENDPOINTS. In other words, we will be evaluating the area A of the region T using three rectangles first and then six rectangles, but in both cases, we’ll be using left hand endpoints.

(b).

The graph below shows the graph of f touching the left endpoints of three approximating rectangles. We see that a = -1, b = 2 and the number of strips n = 3. The width (∆x) of each strip is given by:

Therefore,

∆x

= (b – a)/n

∆x

= (2 – [-1])/3 = 1

∆x = 1. Now, the heights of the three rectangles are:

x0

= a

x1

= a + ∆x

= -1

x2 = a + 2∆x

= =

-1 + 1

=

0

-1 + 2(1) =

1

So, using left hand endpoints, the area A3 of T is given by the sum of the areas of the approximating rectangles:

3

A3

=

∑

lim

n→∞

f(xi-1)∆ x

i=1

which gives

3

A3

=

∑

lim n→∞

i=1

A3

=

[(x1 – 1)3 + 2]∆x

A3

=

[(x0)3 + 2]∆x + [(x1)3 + 2]∆x

A3

=

1[(-1)3 + 2]

A3

=

+ [(x2 – 1)3 + 2]∆x

+ 1[(0 )3 + 2]

1[-1 + 2] +

[(xi-1)3 + 2]∆ x

1[0 + 2]

+ [(x3 – 1)3 + 2]∆x

+ [(x2)3 + 2]∆x

+ 1[(1 )3 + 2] +

1[1 + 2]

=

1+2+3

=

6

So, using 3 rectangles and left endpoints, the area A3 of T is 6. Next, we’ll have to use 6 rectangles.

The next graph is that of f touching the left endpoints of six approximating rectangles. We have a = -1, b = 2 and the number of strips n = 6. The width (∆x) of each strip is given by:

Therefore,

∆x

∆x

= (b – a)/n

∆x

= (2 – [-1])/6

=

½

= ½.

The heights of the six rectangles are:

x0

=

a

=

-1

x1

=

a + ∆x

=

-1 + ½

=-½

x2

=

a + 2∆x

=

-1 + 2(½)

= 0

x3

=

a + 3∆x

=

-1 + 3(½)

= ½

x4

=

a + 4∆x

=

-1 + 4(½)

= 1

x5

=

a + 5∆x

=

-1 + 5(½)

= 3/2

Therefore, the area A4 of the region T is given by the sum of the areas of the approximating rectangles:

A4

which gives

A4

=

=

lim

n→∞

6

∑ f(x

lim

n→∞ 6

i=1

∑ [(x i=1

)∆ x

i-1

)3 + 2]∆ x

i-1

=

[(x1 – 1)3 + 2]∆x + [(x2 – 1)3 + 2]∆x + [(x3 – 1)3 + 2]∆x + [(x4 – 1)3 + 2]∆x + [(x5 – 1)3 + 2]∆x + [(x6 – 1)3 + 2]∆x

=

½[(x0 )3 + 2]

=

½[(- 1)3 + 2] + ½[(-½)3 + 2] + ½[(0)3 + 2] + ½[(½)3 + 2] + ½[(1)3 + 2] + ½[(3/2)3 + 2]

=

½[- 1 + 2] + ½[-1/8 + 2] + ½[0 + 2] + ½[1/8+ 2] + ½[1 + 2] + ½[27/8 + 2]

+ ½[(x1 )3 + 2] + ½[(x2 )3 + 2] + ½[(x3 )3 + 2]

[

=

½ 1 + 15/8 + 2 + 17/8 + 3 + 43/8

=

½[123/8]

=

123/16

+ ½[(x4 )3 + 2] + ½[(x5 )3 + 2]

]

7.6875

=

Therefore, the area of region T is 7.6875.

Summary of this second part: Using 3 rectangles and left endpoints, we found the area of T to be 6. Then, using 6 rectangles and endpoints, the area is 7.6875.

(c).

Part(c) of this exercise is asking us to calculate the area of T using 3 rectangles and then 6 rectangles (but in this case, we’ll be using midpoints).

The next graph shows the graph of f touching the midpoints of three approximating rectangles.

We have a = -1, b = 2 and the number of strips n = 3. Thus, the width (∆x) of each strip is given by:

So,

∆x ∆x

= (b – a)/n = (2 – [-1])/3

∆x

=

1

∆x = 1. These strips divide the region T into 3 subintervals:

[x0, x1], [x1, x2],

and [x2, x3],

We need to know the values of x1, x2 and x3, which are given by:

x1

= a + ∆x

=

-1 + 1

= 0

x2

= a + 2∆x

=

-1 + 2(1)

= 1

x3

= a + 3∆x

=

-1 + 3(1)

= 2

Therefore, the values of the subintervals are

[x0, x1]

=

[- 1, 0]

[x1, x2]

=

[0, 1]

[x2, x3]

=

[1, 2]

Thus, the midpoints (x*i) of these subintervals are:

(x*1)

=

½[x0 + x1]

=

½[-1 + 0]

=

-½

(x*2)

=

½[x1 +x2]

=

½[0 + 1]

=

½

(x*3)

=

½[x2 + x3]

=

½[1 + 2]

=

3/2

Therefore, the area of T (using midpoints) is given by the sum:

3

A5

=

∑ f(x

lim

n→∞

)∆ x

* i

i=1

which gives

3

A5

lim

=

n→∞

∑

[(xi*)3 + 2]∆ x

i=1

Therefore,

[(x*1)3 + 2]∆x + [(x*2)3 + 2]∆x + [(x*3)3 + 2]∆x

A5

=

A5

=

1[(- ½)3 + 2]

A5

=

1[(-1/8) + 2] + 1[(1/8) + 2] + 1[(27/8) + 2]

A5

=

+ 1[(½)3 + 2] + 1[(3/2)3 + 2]

15/8 + 17/8 + 43/8

= 75/8

=

9.375

Therefore, the area of T using 3 rectangles and midpoints is 9.375.

Finally, for this part, we will use midpoints of 6 rectangles.

The figure on the following page is the graph of f touching the midpoints of six approximating rectangles.

We have a = -1, b = 2 and the number of strips n = 6. The width (∆x) of each strip is given by:

Therefore,

∆x

= (b – a)/n

∆x

= (2 – [-1])/6 =

∆x = ½.

½

These 6 strips divide the region T into 6 subintervals:

[x0, x1], [x1, x2], [x2, x3], [x3, x4],

[x4, x5],

[x 5 , x 6 ]

and

We need the values of x0, x1, x2, x3, x4, x5 and x6 , which are given by:

x0

=

a

=

-1

x1

=

a + ∆x

=

-1 + ½

= -½

x2

=

a + 2∆x

=

-1 + 2(½)

=

0

x3

=

a + 3∆x

=

-1 + 3(½)

=

½

x4

=

a + 4∆x

=

-1 + 4(½)

=

1

x5

=

a + 5∆x

=

-1 + 5(½)

= 3/2

x6

=

a + 6∆x

=

-1 + 6(½)

=

Therefore, the subintervals are:

[x0, x1] [x1, x2]

= =

[- 1, - ½] [- ½, 0]

[x2, x3]

=

[0, ½]

[x3, x4]

=

[½, 1]

[x4, x5]

=

[1, 3/2]

[x5, x6]

=

[3/2, 2]

2

Thus, the midpoints (x*i) of these subintervals are:

(x*1)

=

½[x0 + x1]

=

½[- 1 + (- ½)]

=

-¾

(x*2)

=

½[x1 + x2]

=

½[(- ½) + 0]

=

-¼

(x*3)

=

½[x2 + x3]

=

½[ 0 + ½ ]

=

¼

(x*4)

=

½[x3 + x4]

=

½[½ + 1]

=

¾

(x*5)

=

½[x4 + x5]

=

½[1 + (3/2)]

=

5/4

(x*6)

=

½[x5 + x6]

=

½[ (3/2) + 2 ]

=

7/4

Therefore, the area of T(using midpoints) is given by:

6

A6

lim

=

n→∞

∑

f(xi*)∆ x

i=1

which gives

6

A6

lim

=

n→∞

So,

∑ [(x

)3 + 2]∆ x

* i

i=1

=

[(x*1)3 + 2]∆x

=

½[(- ¾)3 + 2] + ½[(- ¼)3 + 2] + ½[(¼)3 + 2] + ½[(¾)3 + 2] + ½[(5/4)3 + 2] + ½[(7/4)3 +

A6

=

½[101/64] + ½[127/64] + ½[129/64] + ½[155/64] + ½[253/64] + ½[471/64]

A6

=

½[101/64 + 127/64 + 129/64 + 155/64 + 253/64 + 471/64]

A6

=

½[1236/64]

A6

+

[(x*2)3 + 2]∆x + [(x*3)3 + 2]∆x

+ [(x*4)3 + 2]∆x + [(x*5)3 + 2]∆x + [(x*6)3 +

2]∆x A6 2]

= 309/32

=

9.65625

Therefore, the area of the region T using 6 rectangles and midpoints is 9.65625.

(d).

So far, we have found six different areas of the region T. Let’s compile the results:

AREAS n 3 6 From the six sketches, we find that:

Right-Hand Endpoints

Left-Hand Endpoints

Sample Points (Midpoints)

15

(A1)

6

(A3)

9.375

12.1875

(A2)

7.6875

(A1) 9.65625

(A5) (A6)

A1 and A2 are overestimates because the actual area of T is much less than A1 and A2, A3 and A4 are underestimates because the actual area of T is more than A3 and A4, This therefore leaves us with A5 and A6, which are both good estimates, but A6 is a much better estimate as

you can see from its graph on page 12.

Thus, the area A6 appears to be the best estimate from the sketches above. In fact, using a scientific or graphing calculator, we find that the exact area of T is 9.75. So, you can see that A6 is the closest value, thus it is the best estimate.

EXAMPLE 3 (a)

Use the definition of area to find an expression for under the curve y = x3 from 0 to 1 as a limit.

(b) The following formula for the sum of the cubes of the first n

integers is given below. Use it to evaluate the

limit in part (a):

13 + 23 + 33 + ......+ n3

=

n(n + 1)

2

2

Solution (a).

The problem is to find the area under y = x3 between x = 0 to x = 1. (This means a = x0 = 0 and b = xn = 1). The question however, does not provide the number of strips or rectangles, so we simply represent the number by n.

So, the width (∆x) of each strip is given by:

∆x ∆x

= (b – a)/n = (1 – 0)/n

∆x

= 1/n

Therefore, the width of each strip is 1/n. The height of each strip is given by:

x0

=

x1

=

a + ∆x

x2

=

x3

=

a

=

0

=

0 + 1/n

= 1/n

a + 2∆x =

0 + 2(1/n)

= 2/n

a + 3∆x =

0 + 3(1/n)

= 3/n

and so on. Therefore, a general formula would be:

xn

a + n∆x

=

= 0 + n(1/n) = n/n

Therefore, the sum of the areas of the approximating rectangles is:

f(x1)∆x + f(x2)∆x

+ f(x3)∆x

+ ....................... + f(xn)∆x

Rn

=

Rn

=

(1/n)3∆x

Rn

=

1/n(1/n)3

Rn

=

1/n[(1/n)3 + (2/n)3 + (3/n)3 +...............................+ (n/n)3]

Rn

=

1/n[(13/n3) + (23/n3) + (33/n3) +.......................... + (n3/n3)]

Rn

=

1/n × 1/n3 [13 + 23 + 33 + ............................ + n3]

+ +

(2/n)3∆x 1/n(2/n)3

(3/n)3∆x

+ +

+ ...................... +

1/n(3/n)3 + ..................... +

(n/n)3∆x 1/n(n/n)3

Rn

=

1/n4 [13 + 23 + 33 +...................... + n3]

According to the definition of area (which is the limit of sums), we have

A

=

A

=

lim Rn

n→∞

which equals

lim 1/n4 [13 + 23 + 33 + .............. + n3]

n→∞

Using Sigma notation, this limit could also be written as

A

lim 1/n

=

4

n→∞

n

∑ i3 i=1

(b).

Observe this limit:

lim 1/n4 [13 + 23 + 33 + .............. + n3]

n→∞ We find that the sum

13 + 23 + 33 + .............. + n3 is a sum of cubes. Recall that the formula for the sum of the first n integers is

13 + 23 + 33 + ......+ n3

n(n + 1)

=

2

2

In other words, we are saying

n

∑ i3

2

n(n + 1)

=

2

i=1 Therefore, the limit

A

=

lim 1/n

4

n→∞

n

∑ i3 i=1

can also be written as

A

=

lim

n→∞

1 n

4

n(n + 1) 2

2

=

lim

n→∞

1

n2 + n

n4

2

2

=

(n2 + n)2

1

lim

n→∞

22

n4

=

n→∞

( n2 + n)2 4n4

n4 + 2n3 + n2 4n4

lim

=

lim

n→∞

Simplifying the expression in the square bracket gives

A

lim [ ¼ + ½ (1/n) + ¼ (1/n2)]

=

n→∞

At this point, we will be calculating limits at infinity. In particular, we know that

lim (1/n) = 0

n→∞

So,

A

=

lim (¼) +

n→∞

lim (½)(1/n)

n→∞

+

lim (¼ )(1/n2)

n→∞

Using limit laws, we obtain:

¼ + [½ × 0] + [¼ × 0]

A

=

A

=

¼ + 0 + 0

A

=

¼

=

0.25

Therefore, the area under the curve y = x3 from x = 0 to x = 1 is 0.25.

In the following section, we take a look at a concept similar to the area problem: distance. Just as the tangents and velocities are related, so are areas and distances.

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