UNIT1
Calculus
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 xaxis. 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+11x6 in the interval [1,3] 2. f(x) = x²4x+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) = (x1)(x2)(x3) 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) = (x1)(x2)(x3) f(x) = x6x²+11x6 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) = 3x²6x+11 suppose x = c, we get f’(c) = 3c²6c+11 by Lagrange’s mean value theorem, f’(c) = = = = 3 now we get, 3c²6c+11 = 3 3c²6c+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 = e1 e1 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. 
Key takeaways
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 (x2).
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 
Key takeaways
+ …….+ + …….. 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).
Example1: Evaluate Sol. Here we notice that it is an indeterminate form of . So that , we can apply L’Hospital rule
Example2: 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
Example3: Evaluate Sol. Let f(x) = , then And = 0
= 0
But if we use L’Hospital rule again, then we get
Example4: 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
Example5: 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,
Example6: 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 Example7: 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, 
Key takeaways
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 Step1: 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. Step2: 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 semivertical 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. 
Key takeaways

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 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 
Key takeaways
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 = 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

Key takeaways
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, Example1: 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.
Example2: find the area under the curves where y = x and y = x + 1, x = 2 and yaxis. 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.
Example2: 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.) =
Example1: find the average value of the function f(x) = x³ over the interval[0,1]. Sol. We know that f( avg.) =
= = = 1 / 4
Example2: 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 =
Example1: 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
Example2: 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 =
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
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
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
A = 2. f( avg.) = 3. Work W = 
References
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 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass