Unit-2

Multivariable Calculus-II

2.1 Improper integrals  

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.

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.         

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-

The p-integral

converges when p>1 and diverges when p

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.

Example-1: 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.

Example-2: 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.

Example-3: 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.

Example-4: 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.

Example-5: 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-

2.2 Beta & Gama function and their properties

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.

 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-

Example(1):  Evaluate 

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  y2  (1 – y ) -1/2dy

= (8/2). B (3, 1/2 )

= 642 /15

BETA FUNCTION MORE PROBLEMS

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.

2.3 Dirichlet’s integral and its applications

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.

Suppose D is the triangular region in xy-plane which is bounded by

, then the double integral over D-

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-

2.4 Application of definite integrals to evaluate surface areas and volume of revolutions

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-

And

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.

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-

 =

Key takeaways-

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

2.     Method of cylindrical shells-

References