Important definitions-

Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a ,  if

Differentiability- A function f(x) is said to be differentiable at x=a if

exists      where ‘a’ belongs to I

Rolle’s theorem-

Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions

1. f(x) is continuous in [a , b]

2. f(x) is differentiable in (a , b)

3. f(a) = f(b)

Then there exists atleast a point point c ϵ (a , b) , where a<b , such that f’(c) = 0

Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-

1.f(x) is a continuous function in[a , b] , from the figure without breaks in between  A & B on y = f(x).

2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.

Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.

Slope of the tangent at P or Q,R, S, will be 0,even the curve y = f(x) decreases or increases, that means f(x) is constant.

Derivative of f(x),

f’(c) = 0

That’s why,     f’(c) = 0

3. The slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.

So that,       f(a) = f(b)

Example: Verify Rolle’s theorem for the function f(x) = x(x+3) in interval [-3, 0].

Sol. First we will differentiate the given function with respect to x, we get

f’(x) = (x²+3x) + (2x + 3)

=

This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.

Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).

Here f(x) satisfies all the conditions of Rolle’s theorem,

Then,

f’(x) = 0  , which gives

= 0

We get,

X = 3 and x = -2

Here  we can see that clearly -3<-2<0 , therefore there exists   -2 ∈ (-3,0) such that

f’(-2) = 0

that means the Rolle’s theorem is true for the given function.

Example: Verify Rolle’s theorem for the given functions below-

1.  f(x) = x³ - 6x²+11x-6  in the interval [1,3]

2. f(x) = x²-4x+8  in the interval [1,3]

Sol. (1)

As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = f(3) = 0

Now we find f’(x) = 0

3x² - 12x +11 = 0

We get, x = 2+ and 2 -

Hence both of them lie in (1,3).

Hence the theorem holds good for the given function in interval [1,3]

(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also,  f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5

Hence f(1) = f(3)

 Now the first derivative of the function,

f’(x) = 0

2x – 4 = 0 , gives

X = 2

We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.

This means that the Rolle’s theorem holds good for the given function and given interval.

Example: Verify the Rolle’s theorem for sin x in the interval []

Sol. Suppose f(x) = sin x

We know that sin x is continuous for all x.

Now , f’(x) = cos x exists for all x in () and

f(0

f(0

thus f(x) satisfies all the conditions of Rolle’s theorem.

Now,

f’(x) = 0 that gives ,    cos x = 0

x =

here we notice that both intervals lie in (.

there exists, c =

so that, f’(c) = 0

the Rolle’s theorem has been verified.

Lagranges’s mean value theorem-

Suppose that f(x) be a function of x such that,

1. if it is continuous in [a , b]

2. if it is differentiable in (a , b)

Then there atleast exists a value cϵ (a , b)

Proof:

Let’s define a function g(x),

g(x) = f(x) – Ax  ………………..(1)

Here A is a constant which is to be determined,

So that,  g(a) = g(b)

now,

g(a) = f(a) – Aa

g(b) = f(b) – Ab

so,

g(a) = g(b),

f(a) – Aa = f(b) – Ab ,

which gives,

A =     …………………..(2)

As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.

And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).

And g(a) = g(b) , because of the choice of A.

Hence g(x) satisfies all the conditions of Rolle’s theorem.

So that,

 There exists a value c  such that a<c<b at which g’(c) = 0

Now, differentiate eq. (1) with respect to x, we get

g’(x) = f’(x) – A

here we know that, x = c,

g’(c) = f’(c) – A

as g’(c) = 0, then

f’(c) – A =0

so that,,

f’(c) = A,

from equation (2) , we get

f’(c) =

hence proved.

Example: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].

Sol. As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).

f(x) = (x-1)(x-2)(x-3)

f(x) =  x-6x²+11x-6

now at x = 0, we get

f(0) = -6  and

at x = 4, we get.

f(4) = 6

diff. the function w.r.t.x , we get

f’(x) = 3x²-6x+11

suppose x = c, we get

f’(c) = 3c²-6c+11

by Lagrange’s mean value theorem,

f’(c) =     =     =   = 3

now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore the given function is verified.

Example: Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].

Sol. We already know that the function which is log x is continuous for all x>0.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

which is exists for all x in (1,e)

so that f(x) is differentiable in (1,e).

by Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

then ,

f’(c) =

we get,

c = e-1

e-1 will always lies between 1 and e .

hence the function is verified by Lagrange’s mean value theorem.

Cauchy’s mean value theorem-

Suppose we have two functions f(x) and g(x) of x, such that,

1. Both functions are continuous in [a,b]

2. Both functions are differentiable in (a,b)

3. g’(x) ≠ 0 for any x ϵ (a,b)

These three exists atleast , x = c ϵ (a,b) , at which

Proof:

Suppose, we define a functions,

h(x) = f(x) – A.g(x) …………………….(1)

So that h(a) = h(b) and A is a constant to be determined.

Now,

h(a) = f(a) – Ag(a)

h(b) = f(b) – A.g(b)

So that,

f(a) – Ag(a) = f(b) – A.g(b),

which gives

A =     …………………………….(2)

Now, h(x) is continuous in [a, b] as RHS of eq. (1) is continuous in [a,b] and h(x) is diff. in (a, b) as RHS of eq. (1) is diff. in (a, b)

Also,

h(a) = h(b)

Therefore all the conditions of Rolle’s theorem are satistfied then there exists a Value x = cϵ (a, b)

So that h’(c) = 0

Differentiate eq.(1) w.r.t. x , we get

h’(x) = f’(x) – A.g’(x)

At x = c

h’(c) = f’(c) – A.g’(c)

0 = f’(c) – A.g’(c)

A =

So that , we get

  where a<c<b

Hence the Cauchy’s mean value theorem is proved.

Example: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cosx  in [ 0 , π/2]

Sol.  It is given tha,

f(x) = sin x and g(x) = cos x

Now,

  f’(x) = cos x and g’(x) = - sin x

We know that both the functions are continuous in [ 0 , π/2] and differentiable in     ( 0 , π/2 )

Also,g’(x) = -sin x  ≠ 0 for all x ϵ( 0 , π/2 )

By Cauchy’s mean value theorem, we get

for some c: 0< c <

That means

which gives,

Cot c = 1

C =

Now we see that lies between 0 and

Example: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]

Sol. We are given, f(x) = x⁴ and g(x) = x

Derivative of these fucntions ,

      f’(x) = 4x³ and g’(x) = 2x

put these values in Cauchy’s formula, we get

2c² =

c² =

c =

now put the values of a = 1 and b = 2 ,we get

c = = = (approx)

Hence the Cauchy’s theorem is verified.

1. Rolle’s theorem-

1. f(x) is continuous in [a , b]

2. f(x) is differentiable in (a , b)

3. f(a) = f(b)

2. Lagranges’s mean value theorem-

1. if it is continuous in [a , b]

2. if it is differentiable in (a , b)

Then there atleast exists a value cϵ (a , b)

f’(c) =

3. Cauchy’s mean value theorem-

1. Both functions are continuous in [a,b]

2. Both functions are differentiable in (a,b)

3. g’(x) ≠ 0 for any x ϵ (a,b)

These three exists atleast , x = c ϵ (a,b) , at which

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + ……..     (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)  

We get-

+ …….

Example: Express the polynomial in powers of (x-2).

Sol.  Here we have,

f(x) =

differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

now using Taylor’s theorem-

+ …….     (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) = 

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12   and f’’’’(2) = 0

now put a = 2 and substitute the above values in equation(1), we get-

Example: Expand sin x in powers of

Sol. Let f(x) = sin x

     Then,

=

By using Taylor’s theorem-

+ …….     (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx  , f’’(x) = - sin x    ,     f’’’(x) = - cos x  and so on.

Putting x = π/2 , we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

from equation (1) put a = and substitute these values, we get-

+ …….

= ………………………..

Example: Expand in powers of (x – 1).

Sol.

Here we have-

Now-

Put these values in Taylor’s theorem-

We get-

Example: By using Maclaurin’s series expand tan x.

Sol.

Let-

Put these values in Maclaurin’s series we get-

Example: Expand by using Maclaurin’s series.

Sol.

Let

Put these values in Maclaurin’s series-

Or

1. Taylor’s Theorem-

+ …….+ + ……..

2.     Maclaurin’s Theorem-

+ …….

Let we have two functions f(x) and g(x) and-

Then-

Is an expression of the form

In that case we can say that f(x)/g(x) is an indeterminate for of the type   at x = a.

Now, Let we have two functions f(x) and g(x) and-

Then-

Is an expression of the form  , in that case we can say that f(x)/g(x) is an indeterminate for of the type   at x = a.

Some other indeterminate forms are

In case we get indeterminate form we apply L’Hospital’s rule.

L’Hospital’s rule for form-

Working steps-

1. Check that the limits f(x)/g(x) is an indeterminate form of type   .

(Note- we can not apply L’Hospital rule if it is not in indeterminate form)

2. Differentiate f and g separately.

3. Find the limits of the derivatives .if the limit is finite , then it is equal to the limit of f(x)/g(x).

Example-1: Evaluate

Sol. Here we notice that it is an indeterminate form of .

So that , we can apply L’Hospital rule-

Example-2: Evaluate .

Sol. Let f(x) = and g(x) = .

Here we see that this is the indeterminate form of 0/0 at x = 0.

Now by using L’Hospital rule, we get-

                                                        =

=

                                                       = = 1

Example: Find the value of a, b if

Solution:

Let

   …

By L – Hospital rule

   …

   …

    … (1)

    …

But

From equation (1)

Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.

If we have f(x) and g(x) are two functions such that 

.

If exist or (∞ , -∞), then

Example-3: Evaluate

Sol. Let f(x) = , then

And

= 0

= 0

But if we use L’Hospital rule again, then we get-

Example-4: Evaluate

Sol. We can see that this is an indeterminate form of type 0/0.

Apply L’Hospital’s rule, we get

But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-

We get

=

L’Hospital’s rule for form-

Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that

If has a finite limit, or if it is , then

Theorem- If we have f(x) and g(x) are two functions such that  .

If exist or (∞ , -∞), then

Example-5: Find   , n>0.

Sol. Let f(x) = log x and g(x) =

These two functions satisfied the theorem that we have discussed above-

So that,

Example-6: Evaluate

Sol. Apply L’Hospital rule as we can see that this is the form of

=

Note- In some cases like above example, we can not apply L’Hospital’s rule.

Other types of indeterminate forms-

Example-7: Evaluate

Sol. Here we find that-

So that this limit is the form of 0.

Now,

Change to obtain the limit-

Now this is the form of 0/0,

Apply L’Hospital’s rule-

Example: Evaluate

Solution:

Let

   … 0o form

Taking log on both sides we get,

  …

   …

By L – Hospital Rule

i.e.

Example: Evaluate

Solution:

Let

   …

Taking log on both sides,

  …

By L – Hospital rule,

i.e.

Example: Evaluate

Solution:

Let

    …

Taking log on both sides, we get

    …

By L – Hospital Rule,

1. L’Hospital’s rule for form-

Working steps-

1. Check that the limits f(x)/g(x) is an indeterminate form of type   .

(Note- we can not apply L’Hospital rule if it is not in indeterminate form)

2. Differentiate f and g separately.

3. Find the limits of the derivatives .if the limit is finite , then it is equal to the limit of f(x)/g(x).

Note- If we have f(x) and g(x) are two functions such that 

.

If exist or (∞ , -∞), then

2.     If we have f(x) and g(x) are two functions such that  .

If exist or (∞ , -∞), then

A function f(x) is said to have a minimum value at x = a if there exists a small number ‘h’, such that f(a) < f(a – h) & f(a + h).

And

A function f(x) is said to have a maximum value at x = a if there exists a small number ‘h’, such that f(a) > f(a – h) & f(a + h).

Note- when we take maximum and minimum value together then it is called extreme value.

The points at which the function gets the extreme value are known as turning points.

Step by step procedure to find maxima and minima-

Step-1: find the first derivative of the function f’(x) and put it equals to zero. Now solve the equation and let its roots be a, b, c, etc.

Step-2: find the second derivative f’’(x) and substitute in it by turns x = a, b, c....

If f’’(a) is negative then f(x) is maximum at x = a.

If f’’(a) is positive then f(x) is minimum at x = a.

Example: Show that sin x (1 + cos x) is a maximum when .

Sol.

Let

Then first derivative-

So that-

Now-

So that-

So that-

Therefore f(x) has a maximum at .

Example: Show that the diameter of the right circular cylinder of greatest curved surface which can be inscribed in a given cone is equal to the radius of the cone.

Sol.

Suppose r is the radius OA of the base and be the semi-vertical angle of the given cone.

Now inscribe a cylinder in it with base radius OL = x

Then the height of the cylinder LP = LA cot

The CSA of cylinder-

Now

Therefore S is maximum when x = r/2.

1. A function f(x) is said to have a minimum value at x = a if there exists a small number ‘h’, such that f(a) < f(a – h) & f(a + h).
2. A function f(x) is said to have a maximum value at x = a if there exists a small number ‘h’, such that f(a) > f(a – h) & f(a + h).

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

There are two types of improper integrals,

 The first one is improper integral of first kind in which one or both limits of integration are infinite it means the interval of integration is not defined.

And second one is improper integral of second kind It is one in which both the limits of integration are finite, but the integrand f (x) is not defined (infinite) or discontinuous at a point between a and b inclusive.

(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-

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-

1. 
2. 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

3.      Integral of a constant function-

4.      Constant multiple property-

5.      Interval union property-

6.      If a < c < b, then

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  

Ex. 3.  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    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  y2  (1 – y ) -1/2dy

= (8/2). B (3, 1/2 )

= 642 /15

BETA FUNCTION MORE PROBLEMS

There is wide range of definite integrals such as, from geometric applications to physical world problems.

we will study about these applications one by one-

(1) Area between the curves-

Suppose we want to find out the area between y = f(x) and y = g(x) on interval [a,b] such that f(x) ≥ g(x). see the figure

Let the area between the two curve is A, then

                                                     A =

Let’s do some examples,

Example-1: find the area under the curves where f(x) =  x+4 and g(x) =  3 – x/2 over the interval [1,4]

Sol. Here limits are given a = 1 , b = 4,

We know that, area under the curve,

A =

= 

=

= 

= (16 – 7/4) =  57/4

So that the area of the region is 57/4 unit square.

Example-2: find the area under the curves where y =  x and y =  x + 1,         x = 2 and y-axis.

Sol. When we draw the graph, curve does not meet, but depend on two vertical lines,

Here boundary is [0 , 2]

Area under the curve,

A =

So that the area under the curve is 3.2092 unit square.

Example-2: Determine the area under the curves y = 2x² + 10 and y = 4x + 16.

Sol. Here we will find out the intersection, which are the boundaries of the curve,

2x² + 10 =  4x + 16

2x² - 4x – 6 = 0

2 (x + 1) ( x – 3 ) = 0

We get, x = -1  and x = 3,

We know that, Area under the curve,

A =

(2) Average function value-

We can use definite integrals to find out the average value of a function.

The average value of a function over the interval [a , b]  is given by the following expression ,

f( avg.) =

Example-1: find the average value of the function f(x) = x³ over the interval[0,1].

Sol.  We know that

f( avg.) =

                                    =      =  = 1 / 4

Example-2: find the average value of f(x) = cosx  over the interval [0 , 𝛑/2]

Sol. We know that,

f( avg.) =

Put the values ,  we get

(3) Work-

This one is the easiest use of definite integral

we know that work done, when a constant force F is applied to move the object over a distance d.

W = Fd

Now suppose that the force at any point x is F(x) , the the work done by to move the object from x = a to x = b is,

W =

Example-1: Find the work done on a spring when we compress it from its natural length of 1m to the 0.75m , if the spring constant k = 16 N / m.

Sol. The force is given by,

F =16x

Here we start to compress it at x = 0 and finish at x= 0.25 from its natural length.

So we take lower limit and upper limit 0 and 0.25 respectively.

We know that,

W =

=

= 0.5 N.m

Example-2: A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm?

Sol.  Here,

F = kx

So that,

1200 = 2k

K = 600 N/cm

In that case,

F = 600x

 We know that,

W =

W = , which gives

W  = 3600 N.cm

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-

                                                  =

1. Area between the curves-

A =

2.     f( avg.) =

3.     Work- W =