Module 2 | unit 2 calculus

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MATHS I

Module 2

Calculus

2.1 Definite Integral as a limit of a sum and Its application in summation of series

When we apply limits in indefinite integrals are called definite integrals.

If an expression is written as , here ‘b’ is called upper limit and ‘a’ is called lower limit.

If f is an increasing or decreasing function on interval [a , b], then

Where

Properties-

1. The definite integral applies only if a<b, but it would be appropriate to include the case a = b and a>b as well, in that case-

If a = b, then

And if a>b, then

2. Integral of a constant function-

3. Constant multiple property-

4. Interval union property-

If a < c < b, then

5. Inequality-

If c and d are constants such that for all x in [a , b], then

c(b – a)

Note- if a function f:[a , b]→R is continuous, then the function ‘f’ is always Integrable.

Example-1: Evaluate.

Sol. Here we notice that f:x→cos x is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

Example-2: Evaluate

Sol. Here is an increasing function on [1 , 2]

So that,

…. (1)

We know that-

And

Then equation (1) becomes-

Note- we can find the definite integral directly as-

Example-3: Evaluate-

Sol.

2.2 Beta and Gamma functions and their properties

A function,

Is called gamma function of n , which can be written as,

Some important results-

Ex.1: Evaluate 0∞ x3/2 e -x dx

Solution: 0∞ x3/2 e -x dx = 0∞ x 5/2-1 e -x dx

= γ (5/2)

= γ (3/2+ 1)

= 3/2 γ (3/2 )

= 3/2. ½ γ (½ )

= 3/2 .½.π

= ¾ π

Ex. 2: Find γ(-½)

Solution: (-½) + 1 = ½
γ(-1/2) = γ(-½ + 1) / (-½)

= - 2 γ (1/2 )

= - 2 π

Ex. 3. Show that

Solution: =

= ) .......................

Ex. 4: Evaluate dx.

Solution: Let dx

Put or ;dx =2t dt .

Ex. 5: Evaluate dx.

Solution : Let dx.

Put or ; 4x dx = dt

Definition : Beta function

Properties of Beta function :

Example(1): Evaluate I =

Solution:

= 2 π/3

Example(2): Evaluate: I = 02 x2 / (2 – x ) . Dx

Solution:

Letting x = 2y, we get

I = (8/2) 01 y 2 (1 – y ) -1/2dy

= (8/2) . B(3 , 1/2 )

= 642 /15

BETA FUNCTION MORE PROBLEMS

Relation between Beta and Gamma functions :

Example(1): Evaluate: I = 0a x4  (a2 – x2 ) . Dx

Solution: Letting x2 = a2 y , we get

I = (a6 / 2) 01 y 3/2 (1 – y )1/2dy

= (a6 / 2) . B(5/2 , 3/2 )

= a6 /3 2

Example(2): Evaluate: I = 02 x (8 – x3 ) . Dx

Solution: Let x3 = 8y

I = (8/3) 01 y-1/3 (1 – y ) 1/3 . Dy

= (8/3) B(2/3 , 4/3 )

= 16 π / ( 9 3 )

Example(3): Prove that

Solution : Let

Put

Example(4): Evaluate

Solution :Let

Put

When

;

Also

Example(5): Show that

Solution :

(

0<p<1)

(by above result)

2.3 Applications of definite integrals to evaluate surface

Area under and between the curves-

Total shaded area will be as follows of the given figure( by using definite integrals)-

Total shaded area =

Example-1: Determine the area enclosed by the curves-

Sol. We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

-3

-2

-1

And

x	-3	0	2
y = 7 - x	10	7	5

We get the following figure by using above two tables-

Area of shaded region =

= ( 12 – 2 – 8/3 ) – (-18 – 9/2 + 9)

= 125/6 square unit

Example-2: Determine the area bounded by three straight lines y = 4 – x, y = 3x and 3y = x

Sol. We get the following figure by using the equations of three straight lines-

y = 4 – x, y = 3x and 3y = x

Area of shaded region-

Example-3: Find the area enclosed by the two functions-

and g(x) = 6 – x

Sol. We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

A =

Therefore the area under the curve is 64/3 square unit.

2.4 Areas and volumes of revolutions

The volume of revolution (V) is obtained by rotating area A through one revolution about the x-axis is given by-

Suppose the curve x = f(y) is rotated about y-axis between the limits y = c and y = d, then the volume generated V, is given by-

Example-1: Find the area enclosed by the curves and if the area is rotated about the x-axis then determine the volume of the solid of revolution.

Sol. We know that, at the point of intersection the coordinates of the curve are equal. So that first we will find the point of intersection-

We get,

x = 0 and x = 2

The curve of the given equations will look like as follows-

Then,

The area of the shaded region will be-

A =

So that the area will be 8/3 square unit.

The volume will be

= (volume produced by revolving – (volume produced by revolving

Method of cylindrical shells-

Let f(x) be a continuous and positive function. Define R as the region bounded above by the graph f(x), below by the x-axis, on the left by the line x = a and on the right x = b, then the volume of the solid of revolution formed by revolving R around the y-axis is given by

Example-2: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].

Sol. The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

Then the volume of the solid will be-

Example-3: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 2x - x² over the interval [0 , 2].

Sol. The graph of the function f(x) = 2x - x² will be-

The volume of the solid is given by-

2.5 Multiple Integral

Double integral –

Before studying about multiple integrals , first let’s go through the definition of definition of definite integrals for function of single variable.

As we know, the integral

Where is belongs to the limit a ≤ x ≤ b

This integral can be written as follows-

Now suppose we have a function f(x , y) of two variables x and y in two dimensional finite region R in xy-plane.

Then the double integration over region R can be evaluated by two successive integration

Evaluation of double integrals-

If A is described as

Then,

]dx

Let do some examples to understand more about double integration-

Example-1: Evaluate , where dA is the small area in xy-plane.

Sol. Let , I =

= 84 sq. Unit.

Which is the required area.

Example-2: Evaluate

Sol. Let us suppose the integral is I,

I =

Put c = 1 – x in I, we get

I =

Suppose , y = ct

Then dy = c

Now we get,

I =

As we know that by beta function,

Which gives,

Now put the value of c, we get

Example-3: Evaluate the following double integral,

Sol. Let ,

I =

On solving the integral, we get

Double Integral over Rectangular and general regions

Consider a function f (x, y) defined in the finite region R of the x-y plane. Divide R into n elementary areas A1, A2,…,An. Let (xr, yr) be any point within the rth elementary are Ar

Fig. 6.1

 f (x, y) dA = f (xr, yr)  A

Evaluation of Double Integral when limits of Integration are given(Cartesian Form).

Ex. 1 : Evaluate ey/x dy dx.

Soln. :

Given : I = ey/x dy dx

Here limits of inner integral are functions of y therefore integrate w.r.t y,

I = dx

I =

= =

ey/x dy dx =

Ex. 2 : Evaluate xy (1 – x –y) dx dy.

Soln. :

Given : I = xy (1 – x –y) dx dy.

Here the limits of inner integration are functions of y therefore first integrate w.r.t y.

I = xdx

Put 1 – x = a (constant for inner integral)

 I = xdx

Put y = at  dy = a dt

y	0	a
t	0	1

 I = xdx

I = xa dx 

I =  x (1 – x) dx =  (x– x4/3) dx

I =

I = =

 xy (1 – x –y) dx dy =

Ex. 3 : Evaluate

Soln. :

Let, I =

Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.

I = dy

I =

= 

Ex. 4: Evaluate e–x2 (1 + y2) x dx dy.

Soln. :

Let I = e–x2 (1 + y2)  x dy = dy e–x2 (1 + y2)  x dy

= dy e– x2 (1 + y2)  dx

= dy  [∵ f (x)  ef(x) dx = ef(x) ]

= (–1) dy (∵ e– = 0)

= = =

 e–x2 (1 + y2) xdx dy =

NOTES:

Type II: Evaluation of Double Integral when region of Integration is provided (Cartesian form)

Ex.1: Evaluate  y dx dy over the area bounded by x= 0 y = and x + y = 2 in the first quadrant

Soln. :

The area bounded by y = x2 (parabola) and x + y = 2 is as shown in Fig.6.2

The point of intersection of y = x2 and x + y = 2.

x + x2 = 2 x2 + x – 2 = 0

 x = 1, – 2

At x = 1, y = 1 and at x = –2, y = 4

Fig. 6.2

 (1, 1) is the point of intersection in Ist quadrant. Take a vertical strip SR, Along SR x constant and y varies from S to R i.e. y = x2 to y = 2 – x.

Now slide strip SR, keeping IIel to y-axis, therefore y constant and x varies from x = 0 to x = 1.

 I =

= (4 – 4x + – ) dx

= = 

 I = 16/15

Ex. 2 : Evaluate  over x  1, y 

Soln. :

1a1 Let I =  over x  1, y 

The region bounded by x  1 and y 

Is as shown in Fig. 6.3.

Fig. 6.3

Take a vertical strip along strip x constant and y varies from y =

To y = . Now slide strip throughout region keeping parallel to y-axis. Therefore y constant and x varies from x = 1 to x = .

I =

= [ ∵ dx = tan–1 (x/a)]

= =

= – = (0 – 1)

 I =

Ex. 3 : Evaluate  (+ ) dx dy through the area enclosed by the curves y = 4x, x + y = 3 and y =0, y = 2.

Soln. :

Let I =  (+ ) dx dy

The area enclosed by the curves y = 4x, x + y =3, y = 0 and y = 2 is as shown in Fig. 6.4.

(find the point of intersection of x + y = 3 and y = 4x)

Fig. 6.4

Take a horizontal strip SR, along SR y constant and x varies from x = to x = 3 – y. Now slide strip keeping IIel to x axis therefore x constant and y varies from y = 0 to y = 2.

I = dy (+ ) dx

= +dy

 I =

= + – 6 + 18

 I =

2.6 Change the order of the integration

Ex.1: Change the order of integration for the integral and evaluate the same with reversed order of integration.

Sol:

Given, I = …(1)

In the given integration, limits are

y = , y = 2a – x and x = 0, x = a

The region bounded by x2 = ay, x + y = 2a Fig.6.5

And x = 0, x = a is as shown in Fig. 6.5

Here we have to change order of Integration. Given the strip is vertical.

Now take horizontal strip SR.

To take total region, Divide region into two parts by taking line y = a.

1 st Region:

Along strip, y constant and x varies from x = 0 to x = 2a – y. Slide strip IIel to x-axis therefore y varies from y = a to y = 2a.

 I1 = dy xy dx …(2)

2nd Region:

Along strip, y constant and x varies from x = 0 to x= . Slide strip IIel to x-axis therefore x-varies from y = 0 to y = a.

 I2 = dy xy dx …(3)

From Equation (1), (2) and (3),

 = dy xy dx + dy xy dx

= + y dy

=dy + (ay) dy = y (4a2 – 4ay + y2) dy + ay2 dy

= (4a2 y – 4ay2 + y3) dy + y2 dy

= +

= a4

Ex. 2 : Evaluate

I =

Soln. :

Given : I = …(1)

In the given integration, limits are

x = 0, x = a, y = 0, y =

The bounded region is as shown in Fig. **.

In the given, strip is vertical. Now take horizontal strip SR. Along strip y constant and x varies from x = 0 to

x = . Slide strip IIel to X-axis therefore y varies from y = 0 to y = a.

 I = dy

Put a2 – y2 = b2

 I =

= =

= dy = dy

= [∵ a = a loge]

 I = dy =

Ex.3 : Express as single integral and evaluate dy dx + dy dx.

Soln. :

Given : I = dy dx + dy dx

I = I1 + I2

The limits of region of integration I1 are
x = – ; x = and y = 0, y = 1 and I2 are x = – 1,
x = 1 and y = 1, y = 3.

The region of integration are as shown in Fig. 6.7

To consider the complete region take a vertical strip SR along the strip y varies from y = x2 to
y = 3 and x varies from x = –1 to x = 1. Fig. 6.7

I =

NOTES:

Evaluation of Double Integral by Changing Cartesian to Polar co-ordinates (when limits are given).

Ex. 1 : Evaluate

Soln. :

The region of integration bounded by

y = 0, y = and x = 0, x = 1

y =  x2 + y2 = x

The region bounded by these is as shown in Fig. 6.8.

Convert the integration in polar co-ordinates by using x = r cos , y = r sin  and dx dy = r dr d

 x2 + y2 = x becomes r = cos 

y = 0 becomes r sin  = 0  = 0

x = 0 becomes r cos  = 0  =

And x = 1 becomes r = sec  Fig. 6.8

Take a radial strip SR with angular thickness , Along strip  constant and r varies from r = 0 to r = cos . Turning strip throughout region therefore  varies from = 0 to  =

 I = r dr d

= 4 cos  sin  d r dr

= 4 cos  sin  d [–]

= – 2 cos  sin  [+1] d

= – 2 [cos  sin  – cos  sin ] d

= –2 + 2 cos  sin  d

= – + 2

= + 1 =

Ex. 2: Sketch the area of double integration and evaluate

dxdy

Soln. :

Let I = dxdy

The region of integration is bounded by the curves

x = y, x = and y = 0, y = Fig. 6.9

i.e. x = y, x2 + y2 = a2 and y = 0, y =

The region bounded by these is as shown in Fig. 6.9.

The point of intersection of x = y and x2 + y2 = a2 is x =

Convert given integration in polar co-ordinates by using polar transformation x = r cos , y = r sin  and dx dy = r dr d

 x = y gives r cos  = r sin  tan  = 1  =

x2 + y2 = a2  r2 = a2 r =  a

y = 0 gives r sin  = 0  = 0.

y = gives r sin  =  r =  cosec 

Take a radial strip SR, along SR  constant and r varies from r = 0 to r = a. Turning this strip throughout region therefore  varies from  = 0 to  =

 I = log r2 r dr d = 2 d r log r dr

I =

= 2 d

= 2 []

I = [/4] =

Evaluation of Double Integral when region of Integration is provided (Polar form)

Ex. 1 : Evaluate  r4 cos3 dr d over the interior of the circle r = 2a cos 

Soln. :

The region of the integration is as shown in Fig. 6.10.Take a radial strip SR, along strip  constant and r varies from r = 0 to r = 2a cos. Now turning this strip throughout region therefore  varies from  = to  =

 I = r4 cos3 dr d

= cos3 d

= cos3 cos5 d

= cos8 d Fig. 6.10

=  2 

I =

 r4 cos3 dr d =

Ex. 2 : Evaluate  r sin  dr d over the cardioid r = a (1 – cos ) above the initial line.

Soln. :

The cardioid r = a (1 – cos ) is as shown in Fig. 6.11. The region of the integration is above the initial line.

Take a radial strip SR, along strip  constant and r Varies from r = 0 to r = a (1 – cos ).

11a New turning the strip throughout region therefore  varies from  = 0 to  = .

 I = r sin  dr d

= sin  d

Fig.6.11

= sin  [a2 (1 – cos )2]

I = (sin  – 2 sin  cos  + sin ) d

=  2 (sin  – sin2  + sin ) d

= +

I = a2= a2

I =

2.7 Applications of multiple integral for calculating area and volumes of the curves.

Area in Cartesian coordinates-

Example-1: Find the area enclosed by two curves using double integration.

y = 2 – x and y² = 2 (2 – x)

Sol. Let,

y = 2 – x ………………..(1)

And y² = 2 (2 – x) ………………..(2)

On solving eq. (1) and (2)

We get the intersection points (2,0) and (0,2) ,

We know that,

Area =

Here we will find the area as below,

Area =

Which gives,

= ( - 4 + 4 /2 ) + 8 / 3 = 2 / 3.

Example-2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.

Sol. Let,

y ² = 4ax ………………..(1)

And

x² = 4ay…………………..(2)

Then if we solve these equations, we get the values of points where these two curves intersect

x varies from y²/4a to and y varies from o to 4a,

Now using the conceot of double integral,

Area =

Area in polar coordinates-

Example-3: Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration.

Sol. We have,

r = a(1+cosθ) …………………….(1)

And

r = a ……………………………….(2)

On solving these equations by eliminating r , we get

a(1+cosθ) = a

(1+cosθ) = 1

Cosθ = 0

Here a θ varies from – π/2 to π/2

Limit of r will be a and 1+cosθ)

Which is the required area.

Example-4: Find the are lying inside a cardioid r = 1 + cos θ and outside the parabola r(1 + cos θ) = 1.

Sol. Let,

r = 1 + cos θ ……………………..(1)

r(1 + cos θ) = 1……………………..(2)

Solving these equations , we get

(1 + cos θ )( 1 + cos θ ) = 1

(1 + cos θ )² = 1

1 + cos θ = 1

Cos θ = 0

θ = ±π / 2

So that, limits of r are,

1 + cos θ and 1 / 1 + cos θ

The area can be founded as below,

Volume by triple integral-

The volume of solid is given by

Volume =

In Spherical polar system

In cylindrical polar system

Ex.1: Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane

Solution: Volume = ………. (1)

Put ,

From equation (1) we have

V =

=24

=24 (u+v+w=1) By Dirichlet’s theorem.

=24

= = = 4

Volume =4

Ex.2: Find volume common to the cylinders, .

Solution: For given cylinders,

, .

Z varies from

Z=- to z =

Y varies from

y= - to y =

x varies from x= -a to x = a

By symmetry,

Required volume= 8 (volume in the first octant)

= 8dx

Volume = 16

Ex.3: Evaluate

1. Solution: -

Ex.4:

Ex.5:Evaluate

Solution:- Put

NOTES:

MASS OF A LAMINA :- If the surface density ρ of a plane lamina is a function of the position of a point of the lamina, then the mass of an elementary area dA is ρ dA and the total mass of the lamina is

In Cartesian coordinates, if ρ = f(x, y) the mass of lamina, M=

In polar coordinates, if ρ = F(r, Ө) the mass of lamina, M=

Ex.1: A lamina is bounded by the curves and . If the density at any point is then find mass of lamina.

Solution:

Ex.2:If the density at any point of a non-uniform circular lamina of radius’ a’ varies as its distance from a fixed point on the circumference of the circle then find the mass of lamina.

Solution:

Take the fixed point on the circumference of the circle as origin and diameter through it as x axis. The polar equation of circle

And density.

Mean Value:

The mean value of the ordinate y of a function over the range to is the limit of mean value of the equidistant ordinates as

Mean Square values of function over the range to is defined as

Mean Square values of function

Root Mean Square Value: (R.M.S. Value):

If y is a periodic function of x of period p, the root mean square value of y is the square root of the mean value of over the range to , c is constant.