Unit2
Multivariable CalculusII
Definite integrals
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.
Improper integrals of first kind
When one or both limits of integration are infinite then this is called improper integral.
Improper integrals
(1) Let f is function defined on [a, ∞) and it is integrable on [a , t] for all t >a, then
If exists, then we define the improper integral of f over [a, ∞) as follows
(2) Let f is function defined on (∞,b] and it is integrable on [t , b] for all t >b, then
If exists, then we define the improper integral of f over (∞ , b] as follows
(3) Let f is function defined on (∞, ∞] and it is integrable on [a , b] for every closed and bounded interval [a , b] which is the subset of R., then
If and exist for some c belongs to R , then we define the improper integral of f over (∞ ,∞ ) as follows
= +
(4) Let f is function defined on (a ,∞) and exists for all t>a , then
If exists , then we define the improper integral of f over (a , ∞) as follows
(5) Let f is function defined on (∞ , b) and exists for all t<b , then
If exists , then we define the improper integral of f over (∞ , b) as follows
Note
Improper integrals over finite intervals
(1) Let f is function defined on (a, b] and exists for all t ∈(a,b) , then
If exists , then we define the improper integral of f over (a , b] as follows
(2) Let f is function defined on [a, b) and exists for all t ∈(a,b) , then
If exists, then we define the improper integral of f over [a , b) as follows
(3) Let f is function defined on [a, c) and (c , b] . if and exist then we define the improper integral of f over [a , b] as follows
We will read more about the uses of improper integrals in the next topic.
Convergence and divergence of the integrals
As we have read about improper integrals, we will understand how to check the convergence and divergence of the integrals.
Rules for convergence/ divergence
(1) If the limit exists and a finite number then the integral is said to be convergent.
(2) If the limit does not exists then the integral is said to be divergent.
Improper integrals of second kind
When the limits of integration are finite but the integrand is not defined at point between a and b.
Let f(x) then the integral has a singularity at the lower limit ‘a’.
Then this singularity is cut off by letting
Where is a very small positive number.
For a convergent improper integral o second kind
Which ignores the contribution of singularity.
Example1: Find out the integral is convergent or divergent. Find the value in case of convergent.
Sol. Here we will convert the integral into limit ,
=
=
=
= ∞
As we can see , here limit does not exist. i.e. that is infinity.
So we can say that the given integral is divergent.
Example2: find out the integral is convergent or divergent. Find the value in case of convergent.
Sol. Covert to the limit ,
=
=
=
Again the limit does not exist that means the integral is divergent.
Example3: Find out the integral is convergent or divergent. Find the value in case of convergent.
Sol. Follow the same process as we did above,
Here limit exists and finite , so that the integral is convergent. And its value is 2√3.
Example4: Find out the integral is convergent or divergent. Find the value in case of convergent.
Sol. As we see, the given is integrand is not continuous at x = 0 , we will split the integral,
= +
We will check one by one whether the integrals are convergent or divergent,
as we found that, integral is divergent
we don’t need to check for the second one.
Example5: Test for convergence or divergence of the following improper integral
Sol. This is a improper integral of first kind and limits are infinite,
Now
Integral convergent.
Key takeaways
Where
2. if a function f:[a , b]→R is continuous, then the function ‘f’ is always Integrable.
3. Let f is function defined on [a, ∞) and it is integrable on [a , t] for all t >a, then
4. If exists, then we define the improper integral of f over [a, ∞) as follows
The beta and gamma functions are defined as
And
These integrals are also known as first and second Eulerian integrals.
Note Beta function is symmetrical with respect to m and n.
Some important results
Ex.1: Evaluate dx
Solution dx = 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
Ex3. Show that
Solution:
=
=
= ) .......................
=
=
Ex. 4: Evaluate dx.
Solution: Let dx
Put or ;dx
=2t dt .
dt
dt
Ex. 5: Evaluate dx.
Solution: Let dx.
x  0  
t  0 
Put or ;
4x dx = dt
dx
Evaluation of beta function 𝛃 (m, n)
Here we have
Or
Again integrate by parts, we get
Repeating the process above, integrating by parts we get
Or
Evaluation of gamma function
Integrating by parts, we take as first function
We get
Replace n by n+1,
Relation between beta and gamma function
We know that
………… (1)
…………………..(2)
Multiply equation (1) by , we get
Integrate both sides with respect to x within limits x = 0 to x = , we get
But
By putting λ = 1 + y and n = m + n
We get by using this result in (2)
So that
Definition : Beta function

Properties of Beta function : 
2. 
3. 
4. 
Example(1): Evaluate
Solution:
= 2 π/3
Example(2): Evaluate: I = 02 x2 / (2 – x ) . dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y2 (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) 
Key takeaways
1.The beta and gamma functions are defined as
And
These integrals are also known as first and second Eulerian integrals.
3. Beta function is symmetrical with respect to m and n.
We use Dirichlet’s integral to find triple and double integrals by expressing in beta and gamma function.
We can find plane area, volume of a solid region, centroid by using Dirichlet’s integral.
2. Suppose V is the solid region, tetrahedron in the first octant, bounded by the plane x + y + z = 0. Then the triple integral over V
Example: Find the area contained in the first quadrant enclosed by the curve given that density at any point p(x, y) is k
Sol. The area A of the plane region is
Now put then x = and then
Where , So that
Dirichlet’s theorem for three variables
Suppose l, m, n are positive, then the triple integral
Where V is the region
Example: Find the mass of an octant of the ellipsoid , the density at any point being p = kxyz
Sol.
As we know that
Put and u+v +w = 1
Therefore
Then mass
Key takeaways
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.
Areas and volumes of revolutions
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
=
Key takeaways
The volume of revolution (V) is obtained by rotating area A through one revolution about the xaxis is given by
2. Method of cylindrical shells
References