Integral Calculus | Mathematics for Competitive Exams PDF Download

Differentiation Under the Integral Sign

In this section, we shall prove that, under suitable conditions, ‘the derivative of the integral and the integral of the derivative are equal’, and consequently, ‘the two repeated integrals are equal for continuous functions.’

Theorem (Leibnitz’s Rule). If f is defined and continuous on the rectangle R = [a, b; c d], and if

(i) f_x(x, y) exists and is continuous on the rectangle R, and

(ii) g(x) =

then g is differentiable on [a, b], and

i.e.,

Corollary 1. (General Leibnitz’s rule). If f satisfy the conditions of the above theorem, and if

(i) θ, ψ : [a, b] → [c, d] are both differentiable, and

(ii)

then g is differentiable on [a, b] and

Corollary 2. If f is continuous on R = [a, b; c, d], then

Iterated Integrals (or repeated integrals)

Definition. An iterated integral is an integral of the form

where θ₁or θ₂ or both are functions of x or constants.

This means that for each fixed x between a and b, the integral

is evaluated, and then the integral

or   ----(1)

The other repeated integral

 ---(2)

is defined in the same way.

Example 1: If I a I < 1. show that

We write

It may be seen that the integrand has only a removable discontinuity at π/2 as much as

Also

We see that f and fa are both continuous functions of two variables in the domain {(a,x) : 0 ≤ x ≤ π, I a I < 1}. If we assign to f, the value a for x = π/2.

We write

⇒    ----(1)

Put

put tan (x/2) = t

diff w. r. to x we get

sec²(x/2).1/2 dx = dt

=

=

=

=

Therefore,

⇒

From this we obtain 

∅(a) = πsin^-1 a + c. . ..( 2 ) 

where, c, is an arbitrary constant.

From (1), we have 

∅(0) = 0.

Putting a = 0 in (2), we obtain c = 0. 

Hence ∅(a) = πsin^-1 a.

Fundamental Theorem of Calculus

Theorem : If a function f is bounded and integrable on [a, b], then the function F defined as

is continuous on [a, b], and furthermore, if f is continuous at a point c of [a, b], then F is derivable at c and

F’(c) = f(c).

Since f is bounded therefore ∃ a number K > 0 such that 

I f(x) I < K, ∀ x ∈ [a, b].

If x₁, x₂ are two points of [a, b] such that a ≤ x₁ < x₂ ≤ b, then

Thus for a given ∈ > 0, we see that

Hence the function F is continuous (in fact uniformly) on [a, b].

Let f be continuous at a point c of [a, b], so that for any ∈ > 0 there exists δ > 0 such that 

Let c - δ < δ < c < t < c + δ

∴

⇒ F'(c) = f(c).

i.e., continuity of f at any point of [a, b] implies derivability of F at that point.

Note. As c is any point of [a, b], we have for all x ∈ [a, b],

F’(t) = f(t) ⇒ F = f

i.e.; continuity of f on [a, b] implies derivability of F on [a, b].

This theorem is sometimes referred to as the First Fundamental Theorem of Integral Calculus.

Definition : A derivable function F, if it exists such that its derivative F’ is equal to a given function f, is called a primitive of f.

The above theorem shows that a sufficient condition for a function to admit of a primitive is that it is continuous. Thus every continuous function f possesses a primitive F where

Remark : We shall now show, with the help of an example, that continuity of a function is not a necessary condition for the existence of a primitive, in other words, “functions possessing primitives are not necessarily continuous”.

Consider the function f on [0, 1], where

It has a primitive F. where

Clearly F’(x) = f(x) but f(x) is not continuous at x = 0, i.e., f(x) is not continuous on [0, 1] In fact, all this amounts to saying that the derivative of a function is not necessarily continuous. 

Theorem : Function f is bounded and integrable on [a, b] and there exists a function F such that F’ = f on [a, b], then

Since the function F'. f is bounded and integrable, therefore for every given ∋ > 0, ∃ δ> 0 such that for every partition P = { a = x₀ x₁  ..x₂..... x_n = b}. with norm μ(p) < δ,

or  ----(1)

for every choice of points t₁ in Δx₁..

Since we have freedom in the selection of points t_t Δx_t, we choose them in a particular way as follows:

By Lagrange’s mean value theorem, we have

=

⇒

= F(b) - F(a).

Hence from (1).

It is sometimes referred to as The second Fundamental Theorem of Integral Calculus.

Example 2 : Show that the function [x], where [x] denotes the greatest integer not greater than x, is integrable in [0, 3].

Since the function is bounded and has only three points of discontinuity therefore it is integrable and

=

1, 2 and 3 being respectively the points of discontinuity of the three integrals on the right.

Example 3 : f is a non-negative continuous function on [a, b] and . Prove that f(x) = 0, for all x e∈ [a, b].

Suppose that, for some c ∈ (a, b), f(c) > 0

Then, for  continuity of f at c implies that, there exists a δ > 0 such that

Now

 

which is a contradiction. Thus f(x) = 0, ∀ x ∈ (a, b). 

Similarly, f(a) > 0, and f(b) > 0. Hence the result follows.

Double Integral

The Calculation of a double integral 

Equivalence of a double with repeated integrals.

Theorem. If the double integral

exists where R is the rectangle [a, b; c, d]

and if also  exist ∀ y ∈ [c, d], 

then the repeated integral

exists and is equal to the double integral.

Definite Double Integral : If we have function f(x, y) then

If dx first then function of y is constant and if dy first then function of x is constant.

Cor. If a double integral exists, then the two repeated integrals cannot exist without being equal. 

Cor. A function f is defined in [0, 1; 0, 1] as-follows :

f(x, y) = 1/2, when y is rational,

f(x, y) = x, when y is irrational;

 exists and is equal to 1/2, but the double integral and the second repeated integral do not exist.

Example 4 : Evaluate  over the area between y = x² and y = x.

The area is bounded by the curves 

y = f₁(x) = x², y = f2(x) = x. see in fig.

When f₁(x) = f₂(x), x² = x, i.e., x = 0, x = 1, i.e., the area of integration is bounded by 

y = x², y = x, x = 0 , x = 1 

In fig. x varies from x = 0 to x = 1 and y varies from y = x² to y = x

∴

=

=

= 3/56

Example 5 : Prove that

LHS =

=

=

=

=

=

=

=

=

RHS =

=

=

=

=

=

=

=

LHS ≠ RHS

Change of Order of Integration

In calculus, interchange of the order of integration is a methodology that transforms iterated integrals of functions into other.

Theorem : If S is a region of Type I, then

If S is a region of Type II, then

We should point out that, unlike the case of integrals over rectangles, there is only one order in which we can carry out the integrations. If S is a Type I region we have to integrate over y before we integrate over x and if S is a Type II region we have to carry out the integration over x before we integrate over y.

However, sometimes a region is both Type I and Type II. This does not mean that

because the right hand side doesn’t really make any sense. Rather, it means that we can prescribe the region S in two different ways

The preceding theorem applied to this situation simple says

In order to reverse the order of integration of an integral like

one therefore first has to Fig. out how to parameterize the region of integration 

as a Type II region; that is to say, one has to Fig. out what c, d, ψ₁(y), and ψ₂(y) are.

Example 6 : Find the value of  if the domain E integration is the triangle bounded by the straight lines y = x, y = 0 and x = 1.

How to take limit of y and x, we draw a strip parallel to y axis in the fig.

Now lower end of strip is on the line y = 0 and upper end of strip is on the line y = x So y varies from y = 0 to y = x.

For x, we pull the strip at the left end of shaded fig i.e. x = 0 and extend the strip i.e the right end of shaded region i.e. x = 1 So y varies from y = 0 to x.

To avoid integration of e^y/x with respect to x, we use

=

=

Example 7 : Evaluate the double integral , when E is the domain which lies between two squares of sides 2 and 4, with center at the origin and sides parallel to the axes.

Domain E is not quadratic with respect to any axes but the straight lines x = -1, x = 1 divide it into four quadratic domains, E₁, E₂, E₃, E₄.

see the region E_1, limit of x is from -2 to -1 and limit of y is from -2 to 2 Similarly find limits for other regions also.

∴

=

=

=  

=

=
=

= 2 cosh(4) - 2 cosh(2)

Example 8 : Evaluate , over the rectangle R = [0, 1; 0, 1], where

For non-zero values of the function f(x, y), the domain R is divided into two domains A and B, we obtain

=

=

=

Example 9 : Evaluate , over the rectangle R = [0, 1 : 0, 2], where [x + y] denotes the greatest integer less than or equal to (x + y). We have, for (x, y) ∈ R.

The domain of integration R (i.e. region ABDE) is divided into three domains ABF, BCEF and CDE and [x + y] = 0 in the region ABF, [x + y] = 1 in region BCEF and [x + y] = 2 in the region CDE.

Therefore, we have

=

=

=

= 3 [1 - 0] + 2 [2 - 0] 

= 3 + 4 = 7

Jacobian of Transformation

f(x, y)dx dy = f(r cosθ, r sinθ) Jdr dθ 

dx dy = Jdr dθ

∴

which is known as Jacobian.

Jacobian of transformation form of (x, y) into (r, θ)

J =

Change of Variables in Double Integrals

Let the functions x = X(u, v); y = Y(u, v) map in one-to-one manner a domain D in Cartesian coordinates (x, y) onto a domain D’ in the new coordinates u, v.

Let f(x, y) = f(X(u, v), Y(u, v)) = F(u, v)

Then

where the Jacobian J =

In case of a transformation from a Cartesian coordinate system (x, y) to polar coordinate system (r, θ); I f I = r, and hence dx dy = r dθ dr.

Change of Variables in Triple Integrals 

Let the functions

x = X(u,v,w), y = Y(u,v,w), z = Z(u,v,w)

map in one-to-one manner, a domain D in cartesian coordinates (x,y,z) onto a domain D’ in the new variables (u,v,w)

Let f(x,y,z) = f(X(u,v,w), Y(u,v,w), Z(u,v,w)) = F(u,v,w)

Then

where the Jacobian J =  (matrix is same as given by eq. (1) of order 3)

The following two transformations because of their frequent occurrence, deserve special mention 

(i) Cylindrical polar coordinates

x = r cosθ, y = r sinθ, z = z Here the Jacobian J = r.

ii) Spherical polar coordinates. x = r sinθ cos ∅, y = r sinθ sin ∅, z = r cosθ 

Here the Jacobian J = r² sinθ

Examples 10 : To integrate (x2 + y2) over the circle x2 + y2 = a², we change to polar, x = r cosθ, y = r sinθ, so that .

 (integrating wrt r)

=

1. To evaluate over the positive quadrant of the ellipse x²/a² + y²/b² = 1, by putting x = au ⇒ d x = a du, and y = bv = > d y = bdv, we have to evaluate ab  dv, over the positive quadrant of the circle u² + v² = 1.

Again, changing to polars, by putting u = r cos 0, v = r sin 0, the integral becomes

=

=

put 1 + r² = p² ⇒ 2rdr = 2pdp (lim it: r = 0 ⇒ p = 1, r = 1⇒ p = √2 )

=

=

put p = √2 sint

dp = √2 cost dt

=

Put p = √2 sin t 

dp = - √2 cost dt

=

=

=

=

=

=

Note : The transformation could be effected in one step by putting x / a = rcosθ, y /b = rsinθ, then I J I = abr.

2. To evaluate ∫∫{ 2a² - 2a( x + y) - ( x² + y²) } dx dy over the circle x² + y. + 2a(x + y) = 2a², transform the origin to ( - a, - a ), by putting x + a = u, y + a = v, so t hat the integral becomes ∫∫( 4a² - u. - v²) dudv, over the circle u² + v² = 4a². Changing to polars , we get 8πa⁴.

 over the circle

x² + y² + 2a(x + y) = 2a² 

x² + 2ax + a.+ y² + 2ay + a² = 4a2 

(x + a)² + (y + a)² = 4a² 

put x + a = u, y + a = v 

So the integral becomes

Changing into polar coordinates

=

=

=

=

= 8πa⁴

Example 11 : Evaluate ∫ ∫y - x ) d x dy, over the region E in the xy-plane bounded by the straight lines y = x - 3 , y = x+1,3y + x = 5,3y + x = 7

It is difficult to evaluate the double integral directly; however a simple change of coordinates reduces the domain of integration into a rectangle with sides parallel to the axis.

Set y - x = u, 3y + x = v

so that

and

The new domain is the rectangle R = [-3, 1; 5, 7] in uv-plane.

∴

=

Example 12 : Evaluate the integral  by passing on to the polar coordinates.

The integral in question is the double integral  dx dy over the region enclosed by the triangle y = 0, y = x, x = 1.

In polar coordinates, the equations of these lines are 0 = 0, 0 = rc/4. r cos 0 = 1, so that the domain of integration is 0 < 0 <π/4, 0 < r < sec 0.

∴

=

Dirichlet's Theorem 

The theorem states that

=

where the integral is extended to all +ve values of the variables x_1, x₂, ..., x_n subject to the condition x₁ + x₂ + ..., x_n ≤ 1.

Liouville's Extension of Dirichlet's Theorem

If x, y, z all positive such that 

h,≤ x + y + z ≤ h₂

Example 13 : Evaluate  where E is the region bounded by x = 0, y = 0, x + y = 1.

Now

=

For this type of integrals, two sets of substitutions are possible, which give an integral with constant limits. We discuss both these here.

First method. Put x + y = u, x = uv 

so that y = u(1 - v) and

The Jacobian vanishes when u = 0, i.e., when x = 0 = y. The origin of the xy-plane corresponds to the whole line u = 0 of the uv-plane, so that the correspondence ceases to be one-to-one. To exclude the origin of the xy-plane, we cut out the region along a line x = h parallel to the y-axis and consider the integral on the remaining domain E,, bounded by the lines 

v = 1, u = 1, uv = h

However, in the limit when h -> 0, this new region degenerates into the square bounded by v = 1, u = 1, v = 0, u = h 

Thus

=

=

=

=

Second method. Put x = u, y = (1 - u)v, so that 

1 - x - y = 1 - u - ( 1 - u ) v = (1 - u)(1 - v )

and

which vanishes for u = 1, so that the point (1, 0) in the xy-plane corresponds to the whole line u = 1 in the uv-plane. However, proceeding as in the first method, the new region becomes the square.

u = 0, v = 0, u = 1, v = 1

Thus

=

=

=

=

Example 14 : Evaluate  dxdy, where E is the region bounded by the co-ordinate  axes and x + y = 1 in the first quadrant.

Taking x - y = u, x + y = v, so that x = 1/2(u+v), y = 1/2(v-u) and the Jacobian is 1/2.

∴  -------(1)

Now

=    ----(2)

Hence, from (1) and (2), the required integral is zero.

Triple Integrals

Let f(x, y, z) be a function of three independent variables x, y and z defined at every point of a three-dimensional region V. Divide the region V into n elementary volumes δV₁f δV₂, δVn and let (x_r, y_r, z_r) be any point inside the rth sub-division δV_r. Find the sum

Then

To extend definition of repeated integrals for triple integrals, consider a function f(x, y, z) and keep x and y constant and integrate with respect to z between limits in general depending upon x and y. This would reduce f(x, y, z) to a function of x and y only. Thus, let

Then in ∅ (x, y) we can keep x constant and integrate with respect to y between limits in general depending upon x this leads to a function of x alone, say

Finally ψ(x) is integrated with respect to x assuming that the limits for x are from a to b.

Thus

=

If we put f(x, y, z) = 1, then the volume

Use of definite triple integral in finding volume of 3D shape : 

If we have a 3D-shape in which two very-very closed point (P, Q) are taken as diagonal of a cuboid.

Then volume of this elementary cuboid inside that 3D shape 

dV = dx dy dz

 taken over region

Example 15 : Evaluate the integral 

Let

=

=

=

=

=

=

=

Example 16 : Evaluate  here V is the volume of the cube bounded by the coordinate planes and the planes x = y = z = a.

Here a column parallel to z-axis is bounded by the planes z = 0 and z = a.

Here the region S above which the volume V stands is the region ist he xy-plane bounded by the lines x = 0, x = a, y = 0, y = a.

Hence the given integral

=

=

=

=

=

=

Example 17 :  where V is the closed region bounded by the cylinder z = 4 - x² and the planes x = 0, y = 0, y = 2 and z = 0.

Here a column parallel to z-axis is bounded by the plane z = 0 and the surface z = 4 - x² of the cylinder.

This cylinder z = 4 - x² meets the z-axis, x = 0, y = 0, at (0, 0, 4) and the x-axis, y = 0, z = 0 at (2, 0, 0) in the given region.

Therefore, it is evident that the limits of integration for z are from 0 to 4 - x², for y from 0 to 2 and for x from 0 to 2.

Here the given integral

=

=

=

=

=

=

=

=

Example 18 : Evaluate taken over the region common to the surfaces x² + y² + z² = a², and x² + y² = ax

The region is bounded, above and below by the surfaces

and its projection on the xy-plane is the circular domain D = x² + y² ≤ ax.

∴

=

Changing to polars, the region D becomes the circle, r = a cosθ

0 ≤ θ ≤ π, 0 ≤ r ≤ a cosθ

=

=

Example 19 : Evaluate  taken over the domain bounded by the cylinder x² + y² = 2ax, and the cone z² = k²(x² + y²).

The domain E is bounded above and below by the surface

  and

and its projection on the xy-plane is the circular domain D, x² + y² ≤ 2ax.

∴

=

Changing to polars,

=

=

=

Change of Variables in A Triple Integral

Let f be continuous in a domain E bounded by a surface S in xyz-space.

Also let

x = x (X, Y, Z), y = y (X, Y, Z); z = z (X, Y, Z).

define three functions of three variables X, Y, Z, defined in a domain E, bounded by a surface S, in XYZ-space.

We suppose that these three functions with values 

x (X, Y, Z), y(X, Y, Z), z(X, Y, Z) 

(i) possess continuous first order partial derivatives at each point of E, and Sr 

(ii) transform E, into E and S1 into S. 

(iii) the transformation is one-one. 

(iv) the Jacobian

does not change sign at any point of E₁, even though it may vanish at some points of S₁ It will then be proved that

=

where x, y, z have to be replaced by their values in terms of X, Y, Z in f(x, y, z) in the integral on the right.

Example 20: Show that  taken over the tetrahedron bounded by the planes, x = 0, y = 0, z = 0,

The given integral is same as

First Method . Let us put x + y + z = u, x + y = uv, x = uvw, i.e., std. substitution 

x = uvw, y = uv(1 - w), z = u(1 - v)

It may be seen that when x, y, z are positive and x + y + z < 1, then each of u, v, w lie between 0 and 1 and conversely. So the given region is fully described when 0 ≤ 1 , 0 ≤ v ≤ 1 , 0 ≤ w ≤ 1 .

Also, them

∴

=

Second Method . Put x = u, y = (1 - u)v, z = (1 - u) (1 - v)w so that 

1 - x - y - z = ( 1 - u)(1 - v)(1 - w)

and

∴

=

Example 21 : Evaluate  taken throughout the domain

Change the variables x, y, z to X, Y, Z where x = aX, y = bY, z = cZ, 

so that ∂(x, y, z)/∂(X, Y, Z) = abc.

Thus

taken throughout X² + Y² + Z² ≤ 1.

Changing X, Y, Z to polar co-ordinates r, θ, φ so that X = r sinθ cosφ, Y = r sinθ sinφ, Z = r cosθ,

We have, since

∂(X, Y, Z)/∂(r, θ, φ) = r² sin θ,

It is easily seen that to describe the whole region.

X² + Y² + Z² ≤ 1.

r varies from 0 to 1; 0 varies from 0 to π ; φ varies from 0 to 2π.

Thus (r, 0, φ) varies in the rectangular parallelepiped.

[0, 1; 0, π; 0, 2π]

∴

SURFACE AREA

Definition : The area of the surface 

x = f(u, v), y = g(u, v), z = h(u, v); (u, v) ∈ D is the double integral

 assuming that the functions f, g, h possess continuous first order partial derivatives in D and at no point of D.

Note : Here, we do not propose to derive the expression as given above by basing the derivation on some elementary notion of surface area. Let the definition as given appear very arbitrary, we outline some considerations to call forth reader’s faith in the same.

I. Let the surface be plane. We take 

x = u, y = v, z = 0 

where (u, v) ranges over a domain D in the XY-plane. In fact, the surface coincides with D in the present case.

We have

∴

=

Thus the definition of surface area as now given agrees with that of areas of plane regions.

II. Consider the surface given by z = h (x, y).

We take it as given by 

z = u, y = v, z = h(u, v) 

where (u, v) = (x, y) ∈ D ⊂ R³. 

We have

∴ surface area =  -----(1)

The reader acquainted with the elements of Differential Geometry knows that 

 are the direction ratios of the normal to the surface at any point (x, y, z) , so that if we suppose δσ to be an element of surface lying on the tangent plane at the point, the projection area δA of δσ on the XY-plane is given by

⇒

=

This suggests, on summation, the truth of (1). 

Note. The reader may easily show that

where E =

Smooth surface.

Definition: A surface given by x = f(u, v), y = g(u, v), z = h(u, v); (u, v) ∈ D is said to be smooth, if f, g, h possess continuous first order partial derivatives at each point of D and at no point there of

 vanish simultaneously.

Example 22: The area of the surface of the paraboloid 2z = x² + y² which lies between the planes z = 0, z = 2 is ______

Here

∴

The projection on the plane z = 2 is x² + y² = 4 or r = 2 which is a circle between θ = 0 and θ = 2π (changing to polars by putting x = r cos θ, y = r sinθ). Hence

=

=

=

Example 23 : Find the area of the surface az = xy that lies inside the cylinder (x² + y²)² = 2a²xy.

We have az = xy.

∴  

Surface area =

= changing into polars .(x² + y² =r² , x = r cosθ , y = r sinθ ) 

Also (x² + y²)² = 2a²xy becomes r² = a² sin 2θ curve is a symmetric about origin and see the loop of the curve lies between θ= 0 to π/2.

∴

=

=

=

 -------(1)

where 

 ------(2)

By equation (i),

Example 24 : Compute the surface area S of the sphere x² + y² + z² = a²

The surface area of the sphere is twice the surface are a of the upper half-sphere 

z =

Now

so that

The domain of integration is the circle x² + y² = a² on the xy-plane. 

Thus by formula (1), we have

On passing to polars, we have

Example 25 : Find the area of that part of the surface of the cylinder x² + y² = a² which is cut out by the cylinder x² + z² = a².

The Fig. shows 1/8th of the desired surface

The equation of the surface has the form

 

so that

The domain of integration is a quarter circle x² + z² = a², x ≥ 0, z≥ 0, on the xz-plane. Thus by formula (3), we have

∴ S = 8a².

Evaluation of Volumes

By Triple Integration

The triple integral

carried throughout a region E in space of three dimensions gives the volume of E.

By Double Integration Let C be the boundary of a region E of the xy-plane and let a cylinder be constructed by lines through the points of C parallel to z-axis. Then the volume of the cylinder enclosed between the surfaces

is   and can be easily seen. given by the double integral

Example 26: The volume enclosed by the surfaces x² + y² = cz, x² + y² = 2ax, z = 0.

The limits of z are 0 to  The limits of y are from  and the limits of x are from 0 to 2a.

The value

=

=

=

Put a - x = a sin θ

=

=

=

Example 27 : The sphere x² + y² + z² = a² is pierced by the cylinder (x² + y²)² = a²(x² - y²); prove that the volume of the sphere that lies inside the cylinder is

Here limits of z are from  and therefore 

the volume =

=

Now the equation of the cylinder is 

(x² + y²)² = a²(x² - y²)

Putting x = r cosθ, y = r sinθ, we get 

r² = a² cos2θ.

∴ The limits of r are from  and limits of 0 are from -π/4 to π/4. 

∴ The required volume

=

=

=

=

=

=