CHAPTER:



CHAPTER: APPLICATION OF INTEGRATION

Contents

1 Plane areas

1.1 Area under a curve as a limit of the sum of areas of rectangles

1.2 Area between two curves

2 Volume of revolution

2.1 Rotation about the x-axis and y-axis

2.2 Rotation about other lines

3. Trapezium rule

1. Using trapezium rule to estimate area

2. Using trapezium rule to estimate volume

4 Miscellaneous Examples

( 1 Plane areas

( 1.1 Area under a curve as a limit of the sum of areas of rectangles

Consider a function f(x) in the interval a ( x ( b. We wish to find the area A of the shaded region.

y

f(x)

A

0 a b x

Suppose the interval [a,b] is divided into n equal intervals [x0,x1], [x1,x2], ... , [xn-1,xn] where x0= a and xn = b.

y f(xn-1) f(x)

f(x0) f(x1)

...

0 a=x0 x1 ... xn-1 xn=b x

Hence, each interval is of length, (x = . On each subinterval, construct a rectangle as shown.

Note that area of 1st rectangle, (A1 = f(x0) (x

area of 2nd rectangle, (A2 = f(x1) (x

...

area of nth rectangle, (An = f(xn(1) (x

[pic]The total area of these n rectangles, Sn = f(x0) (x + f(x1) (x + ... + f(xn(1) (x

Clearly, as the rectangles get smaller, i.e as n ( (, Sn ( A, so A = Sn

={f(x0)( x + f(x1)( x + ... + f(xn(1)( x}

= (x

Recall, =

But area of each rectangle, (A = f(x) (x

( = f(x)

( = f(x) if (x ( 0

( A = ( f(x) dx integrating both sides with respect to x

( (x = ( f(x) dx

Hence, integration is a process of summation. Thus, to find areas, we can use integration.

Area between a curve y = f(x) and the x-axis between limits x = a and x = b

is given by A = a b

Similarly, area between a curve y = f(x) and the y-axis between limits y = a and y = b b

Is given by A = a

Note: For areas between the curve and the x-axis, areas above the x-axis will be

evaluated as positive due f(x) is positive and areas below the x-axis will be evaluated as negative because f(x) is negative. So, when finding areas, you must determine if portions of the curve cut the x-axis. If so, the areas must be evaluated separately and then the numerical values of the areas summed up to give the total area.

Similar consideration occurs for finding areas between a curve and the y-axis.

Example 1.1

Find the area bounded by the curve y = x2 - x + 3, the x-axis and the lines x = 2 and x = 4. [18units2 ]

Solution

y

y = x2 - x + 3

3

0 2 4 x

Example 1.2

Find the area between the curve y = sin x and the x-axis for x ( [- , ]. [2 units2 ]

Solution 1 y = sin x

- x

-1

Example 1.3

Find the area R enclosed by the curve 4y2 = 1 ( , the x-axis, the y-axis,

and the line y = . [units2]

Solution

Example 1.4

Sketch the graph of y = x (x - 1) and hence find the area of the region enclosed by the graph of y = x (x - 1) and y = 0 between x = 0 and x = 3. [ units2 ]

Solution

y y = x(x-1)

0 1 3 x

( 1.2 Area between two curves

The area between two curves, f(x) and g(x), is given by where a and b are the lower and upper boundaries of the area.

y y

f(x) f(x)

g(x) g(x)

a b x a b x

Example 1.5

Find the area of the region bounded by the curve y = 2 + x - x2 and the straight line y = x + 1. [units2 ]

Solution

y

2 y = x + 1

y=2+x-x2

-1 0 1 2 x

Example 1.6

Find the area enclosed by the curves y = 2 + x - x2 and y = 2 - 3x + x2. [2units2 ]

Solution y

y = 2 - 3x + x2

2

-1 0 1 2 x

y = 2 + x - x2

( 2 Volume of revolution

( 2.1 Rotation about the x-axis and y-axis

Let f be a function defined on [a,b]. If the region bounded by y = f(x), the x-axis, the lines x=a and x=b is rotated about the x-axis through 3600, we get the solid shown below.

y y

y = f(x)

a b x

0 a b x

z

Let the volume generated be V.

Imagine that V is cut vertically into many small discs, each with radius y and height (x.

Then V may be considered as a sum of the volumes of these small discs.

Since, in general, volume of a disc is ( r2 h,

Therefore, volume of each disc of V is given by ( y2 (x

So, V (

Hence, when (x gets very small, V =

= (by definition of integration as a summation)

Thus, the volume generated by rotating about the x-axis is given by V = ( y2 dx.

Similarly, the volume generated by rotating about the y-axis is given by V =

d

R x = g(y)

c

0 x

Example 2.1

Find the volume generated by rotating about the x-axis through four right angles the area bounded by the curve y = x3 - 2x2 and the x-axis. []

Solution

y

0 2 x

Let f(x) and g(x) be two functions defined on [a,b] with f(x) ( g(x) ( 0 for each x in [a,b]. Let R be the region bounded by y = f(x), y = g(x) and the lines x = a and x = b. If S is the solid of revolution obtained by rotating the region R about the x-axis through 3600,

then its volume is given by y y =f(x)

( f2(x) dx ( ( g2(x) dx = ( [f2(x) - g2(x)]dx. R

y = g(x)

0 a b x

Example 2.2

A region is bounded by the curve y = 1 - x2 and the line x + y = 1. Show this region clearly on a sketch and find the volume of the solid formed when this region is rotated through 4 right angles about the x-axis. []

Solution

y

1

y=1-x2

x

0 1 x+y=1

Example 2.3

R is the region bounded by the curve y = tan-1 x, the y-axis and the line y = . Show R clearly on a sketch. Find the volume of the solid formed by rotating R about the y-axis through 4 right angles. [( ( units3]

Solution y

y = tan-1x

y =

R

0 x

Example 2.4

A region is bounded by the curve y = and the lines y = 2x, x = 0 and y = 4.

Show this region clearly on a sketch and find the

a) area of this region, and

b) volume of solid formed when region is rotated through 4 right angles about the y-axis.

[a) 1 + 2 ln 2 b) ( ] y

Solution y = 2x

y=4

2

y =

0 x

( 2.2 Rotation about other lines

The volume of solid of revolution about the line, y = h is and

the volume of the solid of revolution about the line, x = k is

where a and b are the lower and upper boundaries of the volume.

Example 2.5

The area bounded by the curve y = x2 + 1, the line y = 2 and the y-axis in the first quadrant is rotated 3600 about the line y = 2. Calculate the volume of the solid of revolution. [( units3 ]

Solution

Example 2.6 (NJC 97/1/17b)

The area R is bounded by the curve y = and lines x = 2 and y = 1. Find the exact value of the volume, V, of the solid obtained when R is rotated through four right angles about the line x = 2.

[( (– 8 ln 2)units3. ]

Solution

( 3 Trapezium Rule

( 3.1 Using Trapezium Rule to Estimate Area

When a function f(x) is difficult to integrate, it may be possible to estimate the value of by numerical methods. The trapezium rule enables us to obtain a good approximation to the value . The integral is represented by the area under the curve y = f(x) between x = a and x = b. Suppose that the area is divided into n equal stripes of width H. y0, y1, y2, … yn represent the values of f(x) at x0=a, x1 = a + H, x2 = a + 2H, …, xn = b.

y

y=f(x)

(H( (H( (H(

0 a =x0 x1 x2 x n – 1 x n = b x

An approximate value for the area is obtained by regarding each strip as a trapezium, so that

( H(y0 + y1) + H(y1 + y2) + … + H(yn-1 + yn) ( H(y0 + 2y1 + 2y2 + … + 2yn-1 + yn)

Thus we have the trapezium rule for n equal strips (or n+1 ordinates) as below:

( H[(y0 + yn) + 2(y1 + y2 + … + yn-1 )] where H = .

Note: Whether using the trapezium rule will over estimate or under estimate the area depends on the curve, i.e. whether the actual area under the curve is smaller or bigger than the total areas of the trapeziums.

Also, n ordinates = n ( 1 intervals/trapezia.

Example 3.1

Use the trapezium rule with 5 ordinates to find an approximate value for . [1.104 ]

Note: Check answer by integration: = [tan-1 x ]20 = tan-12 - tan-10 = 1.10715 (to 5 decimal places).

Solution

Example 3.2 (AJC 96/1/9)

Evaluate by the trapezium rule using 5 ordinates, giving your answer correct to 3 significant figures. Give a reason whether your answer is an over or under estimation of the actual value.

[0.619, under]

Solution

( 3.2 Using Trapezium Rule to Estimate Volume

Trapezium rule can also be used to estimate volume. The formula is the same except that it is now applied with the volume formula, i.e. use the appropriate volume formula (y2 dx or (x2 dy, then use the trapezium rule.

Example 3.3 (VJC 00/1/19iv modified)

The diagram shows the sketch of y = , |x | < 1. R is the shaded region. Use the trapezium rule with three ordinates to estimate the volume generated when R is rotated about the x-axis through 2( radians. [1.37]

Solution

1

-1 0 1

Example 3.4 (NJC 99/1/18iv modified)

The shaded region R is bounded by the curves y = and y = , the x-axis, the y-axis and the line

x = . Using the trapezium rule with four ordinates, obtain an approximation to the volume formed when R is rotated through 3600 about the x-axis. [ units3 ]

Solution

y = y =

0

( 4 Miscellaneous Examples

Example 4.1 (YJC 01/1/10)

Using the trapezium rule with 4 trapezia of equal width, find an approximation for , where f(x) = (,

giving your answer correct to 3 decimal places. Hence, estimate (with brief explanation) the value of

, giving your answer correct to 2 decimal places. [1.486, 2.97]

Solution

SUMMARY (Application of Integration)

Area between a curve y = f(x) and the x-axis between limits x = a and x = b

is given by A = a b

Similarly, area between a curve y = f(x) and the y-axis between limits y = a and y = b b

is given by A = a

The area between two curves, f(x) and g(x), is given by where a and b are the lower and upper boundaries of the area.

The volume generated by rotating about the x-axis is given by V = ( y2 dx.

The volume generated by rotating about the y-axis is given by V =

Let f(x) and g(x) be two functions defined on [a,b] with f(x) ( g(x) ( 0 for each x in [a,b]. Let R be the region bounded by y = f(x), y = g(x) and the lines x=a and x=b. If S is the solid of revolution obtained by rotating the region R about the x-axis through 3600,

then its volume is given by y y =f(x)

( [f2(x) - g2(x)]dx. R

y = g(x)

0 a b x

The volume of solid of revolution about the line, y = h is and

the volume of the solid of revolution about the line, x = k is

where a and b are the lower and upper boundaries of the volume.

Trapezium Rule

The trapezium rule for n equal strips (or n+1 ordinates) is as below:

( H{(y0 + yn) + 2(y1 + y2 + … + yn-1)} where H = .

Note: Whether using the trapezium rule will over estimate or under estimate the area depends on the curve, i.e. whether the actual area under the curve is smaller or bigger than the total areas of the trapeziums.

“Do not allow yourself to be disheartened by any failure

as long as you have done your best.” Mother Teresa

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4y2 = 1 (

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