Module 2
Calculus
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.
Example1: 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
Example2: 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
Example3: Evaluate
Sol.
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 5/21 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
X  0  
t  0 
Put or ;dx =2t dt .
dt
dt
Ex. 5: Evaluate dx.
Solution : Let dx.
x  0  
t  0 
Put or ; 4x dx = dt
dx
Definition : Beta function

Properties of Beta function : 
2. 
3. 
4. 
Example(1): Evaluate I =
Solution:
= 2 π/3
Example(2): Evaluate: I = 02 x2 / (2 – x ) . Dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y 2 (1 – y ) 1/2dy
= (8/2) . B(3 , 1/2 )
= 642 /15
BETA FUNCTION MORE PROBLEMS
Relation between Beta and Gamma functions :
 
Example(1): Evaluate: I = 0a x4 (a2 – x2 ) . Dx Solution: Letting x2 = a2 y , we get I = (a6 / 2) 01 y 3/2 (1 – y )1/2dy = (a6 / 2) . B(5/2 , 3/2 ) = a6 /3 2 Example(2): Evaluate: I = 02 x (8 – x3 ) . Dx Solution: Let x3 = 8y I = (8/3) 01 y1/3 (1 – y ) 1/3 . Dy
= (8/3) B(2/3 , 4/3 ) = 16 π / ( 9 3 ) Example(3): Prove that Solution : Let Put or ,Example(4): Evaluate Solution :Let Put or ,,When ,;,
Also
Example(5): Show that Solution : = (0<p<1)(by above result) 
Area under and between the curves
Total shaded area will be as follows of the given figure( by using definite integrals)
Total shaded area =
Example1: 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
x  3  2  1  0  1  2 
1  10  5  2  1  2  5 
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
Example2: 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
Example3: 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.
The volume of revolution (V) is obtained by rotating area A through one revolution about the xaxis is given by
Suppose the curve x = f(y) is rotated about yaxis between the limits y = c and y = d, then the volume generated V, is given by
Example1: Find the area enclosed by the curves and if the area is rotated about the xaxis 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 xaxis, 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 yaxis is given by
Example2: Find the volume of the solid of revolution formed by revolving R around yaxis 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 yaxis over the interval [1 , 3]
Then the volume of the solid will be
Example3: Find the volume of the solid of revolution formed by revolving R around yaxis 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
=
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 xyplane.
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
Example1: Evaluate , where dA is the small area in xyplane.
Sol. Let , I =
=
=
=
= 84 sq. Unit.
Which is the required area.
Example2: 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 =
I =
I =
I =
I =
As we know that by beta function,
Which gives,
Now put the value of c, we get
Example3: 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 xy 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 = 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 yaxis, 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. :
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 yaxis. 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 =
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 xaxis 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 xaxis therefore xvaries 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 Xaxis 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 coordinates (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 coordinates 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 coordinates 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 d
= 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 ).
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 =
Area in Cartesian coordinates
Example1: 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.
Example2: 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
Example3: 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.
Example4: 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 coordinates 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)
=8
=8
= 8dx
=8
=8
=8
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 nonuniform 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
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.
Ex. 1: Find mean value and R.M.S. Value of the ordinate of cycloid
, over the range to .
Sol Let P(x,y) be any point on the cycloid . Its ordinate is y.
=
=
=
R.M.S.Value =
Ex. 2: Find The Mean Value of Over the positive octant of the
Ellipsoid
Sol: M.V.=
Since
Put
M.V.=
References
1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition,Pearson, Reprint, 2002.
2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.
3. Veerarajan T., Engineering Mathematics for first year, Tata McGrawHill, New Delhi, 2008.
4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11thReprint, 2010.
5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.
6. N.P. Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint, 2008.
7. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010.