Table of contents  
Resolution of Vectors  
Procedure to solve the Vector Equation  
Addition and Subtraction in Component Form  
Multiplication of Vectors (The Scalar and vector products) 
If and be any two nonzero vectors in a plane with different directions and be another vector in the same plane. can be expressed as a sum of two vectorsone obtained by multiplying by a real number and the other obtained by multiplying by another real number.
(where l and m are real numbers)
We say that has been resolved into two component vectors namely
(where l and m are real number)
We say that has been resolved into two component vectors namely
and along and respectively. Hence one can resolve a given vector into two component vectors along a set of two vectors  all the three lie in the same plane.
Resolution along rectangular component
It is convenient to resolve a general vector along axes of a rectangular coordinate system using vectors of unit magnitude, which we call as unit vectors. are unit along x, y and zaxis as shown in figure below :
Resolution in two Dimensions
Consider a vector that lies in xy plane as shown in figure,
⇒
The quantities A_{x} and A_{y} are called xand ycomponents of the vector .
A_{x} is itself not a vector but is a vector and so it .
A_{x} = A cosθ and A_{y} = A sinθ
It's clear from above equation that a component of a vector can be positive, negative or zero depending on the value of q. A vector can be specified in a plane by two ways :
(a) its magnitude A and the direction q it makes with the xaxis; or
(b) its components A_{x} and A_{y} A = , θ =
Note : If A = A_{x} ⇒ A_{y} = 0 and if A = A_{y} ⇒ A_{x} = 0 i.e., components of a vector perpendicular to itself is always zero. The rectangular components of each vector and those of the sum are shown in figure.
We saw that
is equivalent to both
C_{x} = A_{x} + B_{x}
and C_{y} = A_{y} + B_{y}
Refer figure (b)
Vector has been resolved in two axes x and y not perpendicular to each other. Applying sine law in the triangle shown, we have
or R_{x} = and R_{y} =
If α+β = 90°, R_{x} = R sinβ and R_{y} = R sin
Ex.7 Resolve the vector along an perpendicular to the line which make angle 60° with xaxis.
Sol.
so the component along line = A_{y} cos 30° + A_{x} cos 60° and perpendicular to line = A_{x} sin 60°  A_{y} sin 30°
Ex.8 Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at 30° to the horizontal
Sol. Component perpendicular to the plane
= = N
and component parallel to the plane
W_{} =W sin 30° = (10) = 5 N
Ex.9 Resolve horizontally and vertically a force F = 8 N which makes an angle of 45° with the horizontal.
Sol. Horizontal component of is
F_{H} = F cos 45° = (8) =
and vertical component of is
F_{v} = F sin 45° = = Ans.
...(1)
(a) There are 6 variables in this equation which are the following :
(1) Magnitude of and its direction
(2) Magnitude of and its direction
(3) Magnitude of and its direction.
(b) We can solve this equation if we know the value of 4 variables [Note : two of them must be directions]
(c) If we know the two direction of any two vectors then we will put them on the same side and other on the different side.
For example
If we know the directions of and and direction is unknown then we make equation as follows:
(d) Then we make vector diagram according to the equation and resolve the vectors to know the unknown values.
Ex.10 Find the net displacement of a particle from its starting point if it undergoes two successive displacements given by , 37° North of West, , 53° North of East
Sol.
Angle from west  east axis (x  axis)
Ex.11 Find magnitude of and direction of . If makes angle 37° and makes 53° with x axis and has magnitude equal to 10 and has 5. (given )
Sol.
B = 5 (magnitude can not be negative) & Angle made by A
Ex.12 Find the magnitude of F_{1} and F_{2}. If F_{1}, F_{2} make angle 30° and 45° with F_{3} and magnitude of F_{3} is 10 N. (given = )
Sol.
Short Method
If their are two vectors and their resultant make an angle α with . then A sin α = β sin β
Means component of perpendicular to resultant is equal in magnitude to the component of perpendicular to resultant.
Ex.13 If two vectors and make angle 30° and 45° with their resultant and has magnitude equal to 10, then find magnitude of .
Sol. B sin 60° = A sin 30°
⇒ 10 sin 60° = A sin 30°
⇒ A =
Ex.14 If and have angle between them equals to 60° and their resultant make, angle 45° with and have magnitude equal to 10. Then Find magnitude of .
Sol. here a = 45° and b = 60° 45° = 15°
so A sinα = B sinβ
10 sin 45° = B sin 45°
So B =
=
Suppose there are two vectors in component form. Then the addition and subtraction between these two are
Also if we are having a third vector present in component form and this vector is added or subtracted from the addition or subtraction of above two vectors then
Note : Modulus of vector A is given by
Ex.15 Obtain the magnitude of if
and
Sol.
Magnitude of
= Ans.
Ex.16 Find and if make angle 37° with positive xaxis and make angle 53° with negative xaxis as shown and magnitude of is 5 and of B is 10.
Sol.
for
+ =
so the magnitude of resultant will be = =
and have angle θ = from negative x  axis towards up
for
So the magnitude of resultant will be
=
and have angle from positive xaxis towards down.
The scalar product or dot product of any two vectors and , denoted as . (read dot ) is defined as the product of their magnitude with cosine of angle between them.
Thus,
(here θ is the angle between the vectors)
Properties :
Geometrically, B cosθ is the projection of onto and vice versa
Component of along = B cosθ = = (Projection of on )
Component of along = A cosθ = = (Projection of on )
i.e., vectors are parallel ⇒
= AA cosθ = A^{2} ⇒
Ex.17 If the vectors and are perpendicular to each other. Find the value of a?
Sol. If vectors and are perpendicular
⇒ ⇒
⇒ a^{2} 2a 3 = 0 ⇒ a^{2} 3a a 3 = 0
⇒ a(a 3) +1 (a 3 ) ⇒ a = 1, 3
Ex.18 Find the component of along ?
Sol. Component of along is given by hence required component
=
Ex.19 Find angle between and ?
Sol. We have cosθ =
cosθ = = θ = cos^{1}
Ex.20 (i) For what value of m the vector is perpendicular to
(ii) Find the component of vector along the direction of ?
Sol.
(i) m = 10 (ii)
Important Note :
Components of b along and perpendicular to a.
Let . represent two (nonzero) given vectors a, b respectively. Draw BM perpendicular to
From ΔOMB, =
Thus are components of b along a and perpendicular to a.
Now
Hence, components of b along a perpendicular to a are.
(a . b/ a^{2}) a and b  (a . b / a^{2}) a respectively.
Ex.21 The velocity of a particle is given by . Find the vector component of its velocity parallel to the line .
Sol. Component of along
The vector product or cross product of any two vectors and , denoted as
(read cross ) is defined as :
Here θ is the angle between the vectors and the direction is given by the right  hand  thumb rule.
Right  Hand  Thumb Rule :
To find the direction of , draw the two vectors and with both the tails coinciding. Now place your stretched right palm perpendicular to the plane of and in such a way that the fingers are along the vector and when the fingers are closed they go towards . The direction of the thumb gives the direction of .
Properties :
Ex.22 is East wards and is downwards. Find the direction of × ?
Sol. Applying right hand thumb rule we find that is along North.
Ex.23 If , find angle between and
Sol. AB cosθ = AB sinθ tanθ = 1 ⇒ θ = 45°
Ex.24 ⇒ here is perpendicular to both and
Ex.25 Find if and
Sol. = =
Ex.26 (i) is NorthEast and is down wards, find the direction of
(ii) Find × if and
Ans. (i) North  West. (ii)
Change in position vector of particle is known as displacement vector.
Thus we can represent a vector in space starting from (x , yj & ending at
CALCULUS
14. Constants : They are fixed real number which value does not change
Ex. 3, e, a, 1, etc.
15. Variable :
Something that is likely to vary, something that is subject to variation.
or
A quantity that can assume any of a set of value.
Types of variables.
(i) Independent variables : Independent variables is typically the variable being manipulated or changed
(ii) dependent variables : The dependent variables is the object result of the independent variable being manipulated.
Ex. y = x^{2}
here y is dependent variable and x is independent variable
16. FUNCTION :
Function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent variable.
The temperatures at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent & temperature is the dependent variable.
The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent and interest is the dependent variable.
In each case, the value of one variable quantity (dependent variable), which we might call y, depends on the value of another variable quantity (independent variable), which we might call x. Since the value of y is completely determined by the value of x, we say that y is a function of x and represent it mathematically as y = f(x).
all possible values of independent variables (x) are called domain of function.
all possible values of dependent variable (y) are called Range of function.
Think of function f as a kind machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (figure).
When we study circles, we usually call the area A and the radius r. Since area depends on radius, we say that A is a function of r, A = f(r). The equation A = πr^{2} is a rule that tells how to calculate a unique (single) output value of A for each possible input value of the radius r.
A = f(x) = πr^{2}. (Here the rule of relationship which describes the function may be described as square & multiply by π)
if r = 1 A = π
if r = 2 A = 4π
if r = 3 A = 9π
The set of all possible input values for the radius is called the domain of the function. The set of all output values of the area is the range of the function.
We usually denote functions in one of the two ways :
1. By giving a formula such as y = x^{2} that uses a dependent variable y to denote the value of the function.
2. By giving a formula such as f(x) =x^{2} that defines a functions symbols f to name the function.
Strictly speaking, we should call the function f and not f(x).
y = sin x. Here the function is y since, x is the independent variable.
Ex.27 The volume V of ball (solid sphere) of radius r is given by the function V(r) =
The volume of a ball of radius 3m is?
Sol. V(3) = = 36 pm^{3}.
Ex.28 Suppose that the function F is defined for all real numbers r by the formula.
F(r) = 2 (r 1) +3.
Evaluate F at the input values 0, 2 x 2, and F(2).
Sol. In each case we substitute the given input value for r into the formula for F:
F(0) = 2(0 1) + 3 = 2 + 3 = 1
F(2) = 2(2 1) + 3 = 2 + 3 =5
F(x + 2) = 2 (x + 2 1) + 3 = 2x + 5
F(F(2)) = F(5) = 2(5 1) 3 = 11
Ex. 29 function f(x) is defined as f(x) = x^{2} + 3, Find f(0), f(l), f(x>), f(x + 1) and f(f(l))
Sol. f(0) = 0^{2} + 3 = 3
f(1) = l2 + 3 = 4
f(x^{2}) = (x^{2})^{2} +3 = x^{4} + 4
f(x +1) = (x + 1)^{2} + 3 = x^{2} + 2x + 4
= f(4) = 42+3 = 19
17. Differentiation
Finite difference :
The finite difference between two values of a physical is represented by Δ notation.
For example :
Difference in two values of y is written as Δy as given in the table below.
y_{2}  100  100  100 
y_{1}  50  99  99.5 
Δy = y_{2}  y_{1}  50  1  0.5 
Infinitely small difference :
The infinitely small difference means veryvery small difference. And this difference is represented by 'd' notation instead of 'D'.
For example infinitely small difference in the values of y is written as 'dy'
if y_{2} = 100 and y_{1} = 99.9999999999999.....
then dy = 0.00000000000000..........00001
Definition of differentiation
Another name of differentiation is derivative. Suppose y is a function of x or y = f(x)
Differentiation of y with respect to x is denoted by symbols f' (x)
where f'(x) = ; dx is very small change in x and dy is corresponding very small change in y.
Notation : There are many ways to denote the derivative of function y = f(x), the most common notations are these :
y  "y prime"  Nice and brief and does not name the independent variable 
dy/dx  " dy by dx"  Names the variables and uses d for derisive 
df/dx  " df by dx"  Emphasizes the function's name 
” d by dx of f"  Emphasizes the idea that differentiation is an operation performed on f.  
D_{x}f  " dx of f"  A common operator notation 
” y dot"  One of Newton's notations, now common for time derivative i.e. dy/dt 
Average rates of change :
Given an arbitrary function y = f(x) we calculate the average rate of change of y with respect to x over the interval (x, x +Δx) by dividing the change in value of y, i.e., Dy = f(x+Δx) f(x), by length of interval Δx over which the change occurred.
The average rate of change of y with respect to x over the interval [x, x+Δx]
Geometrically
= tanθ = Slope of the line PQ
In triangle QPR tanθ =
therefore we can say that average rate of change of y with respect to x is equal to slope of the line joining P & Q.
The derivative of a function
We know that Average rate of change of y w.r.t x is 
If the limit of this ratio exists as Δx → 0, then it is called the derivative of given function f(x) and is denoted as
18. GEOMETRICAL MEANING OF DIFFERENTIATION :
The geometrical meaning of differentiation is very much useful in the analysis of graphs in physics. To understand the geometrical meaning of derivatives we should have knowledge of secant and tangent to a curve.
Secant and Tangent to a Curve
Secant :  A secant to a curve is a straight line, which intersects the curve at any two points.
Tangent :
A tangent is a straight line, which touches the curve a particular point. Tangent is limiting case of secant which intersects the curve at two overlapping points.
In the figure  1 shown, if value of Δx has gradually reduced then the point Q will move nearer to the point P. If the process is continuously repeated (Figure2) value of Δx will be infinitely small and secant PQ to the given curve will become a tangent at point P.
Therefore
we can say that differentiation of y with respect to x, i.e. is equal to slope of the tangent at point P (x,y)
or tanθ =
(From fig1 the average rate change of y from x to x+Δx is identical with the slope of secant PQ)
Rule No. 1 Derivative Of A Constant
The first rule of differentiation is that the derivative of every constant function is zero.
If c is constant, then
Ex.30, ,
Rule No.2 Power Rule
If n is a real number, then
To apply the power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.
Ex.31
Function defined for x > 0 derivative defined only for x > 0
Function defined for x > 0 derivative not defined at x = 0
Rule No.3 The Constant Multiple Rule
If u is a differentiable function of x, and c is a constant, then
In particular, if n is a positive integer, then
Ex.34 The derivative formula
says that if we rescale the graph of y = x^{2} by multiplying each ycoordinate by 3, then we multiply the slope at each point by 3.
Ex.35 A useful special case
The derivative of the negative of a differentiable function is the negative of the function's derivative. Rule 3 with c = 1 gives.
Rule No.4 The Sum Rule
The derivative of the sum of two differentiable functions is the sum of their derivatives.
If u and v are differentiable functions of x, then their sum u+v is differentiable at every point where u and v are both differentiable functions in their derivatives.
The sum Rule also extends to sums of more than two functions, as long as there are only finite functions in the sum. If u_{1}, u_{2}, ........ u_{n} is differentiable at x, then so if u_{1}+u_{2} ....... +u_{n}, then
Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in above example.
Rule No. 5 The Product Rule
If u and v are differentiable at x, then if their product uv is considered, then .
The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation
(uv)' = uv' + vu'.
While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance,
while , which is wrong
Ex.37 Find the derivatives of y = (x^{2}+1) (x^{3}+3)
Sol. Using the product Rule with u = x^{2}+1 and v = x^{3}+3, we find
= (x^{2}+1) (3x^{2}) + (x^{3}+3) (2x)
= 3x^{4} + 3x^{2} + 2x^{4} + 6x = 5x^{4} + 3x^{2} + 6x
Example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check :
y = (x^{2} + 1) (x^{3} + 3) = x^{5} + x^{3} + 3x^{2} + 3
= 5x^{4} + 3x^{2} + 6x
This is in agreement with our first calculation.
There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with.
Ex.38 Let y = uv be the product of the functions u and v. Find y'(2) if u(2) = 3, u'(2) = 4, v(2) = 1, and v'(2) = 2.
Sol.
From the Product Rule, in the form y' = (uv)' = uv' + vu',
we have y'(2) = u(2) v'(2) + v(2) u'(2)
= (3) (2) + (1) (4) = 64 = 2
Rule No.6 The Quotient Rule
If u and v are differentiable at x, and v(x) ¹ 0, then the quotient u/v is differentiable at x,
and
Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives.
Ex.39 Find the derivative of
Sol. We apply the Quotient Rule with u = t^{2} 1 and v = t^{2} 1
Rule No. 7 Derivative Of Sine Function
Ex.40
Rules No.8 Derivative Of Cosine Function
Ex.41 (a) y = 5x + cos x Sum Rule
Product Rule
= sin x(— sin x) + cos x (cos x)
= cos^{2} x  sin^{2} x  cos 2x
Rule No. 9 Derivatives Of Other Trigonometric Functions
Because sin x and cos x are differentiable functions of x, the related functions
;
;
are differentiable at every value of x at which they are defined. There derivatives, Calculated from the Quotient Rule, are given by the following formulas.
;
;
Ex.42 Find dy / dx if y = tan x.
Sol.
Ex. 43
Rule No. 10 Derivative Of Logarithm And Exponential Functions
,
Ex.44 y = e^{x} . log_{e} (x)
⇒
Rule No. 11 Chain Rule Or `Outside Inside' Rule
It sometimes helps to think about the Chain Rule the following way. If y = f(g(x)),
= f'[g(x)] . g'(x)
In words : To find dy/dx, differentiate the "outside" function f and leave the "inside" g(x) alone; then multiply by the derivative of the inside.
We now know how to differentiate sin x and x^{2} 4, but how do we differentiate a composite like sin(x^{2} 4)?
The answer is, with the Chain Rule, which says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points. The Chain Rule is probably the most widely used differentiation rule in mathematics. This section describes the rule and how to use it. We begin with examples.
Ex.45 The function y = 6x 10 = 2(3x 5) is the composite of the functions y = 2u and u = 3x 5. How are the derivatives of these three functions related ?
Sol. We have , ,
Since 6 = 2 × 3
Is it an accident that ?
If we think of the derivative as a rate of change, our intuitions allows us to see that this relationship is reasonable. For y = f(u) and u = g(x), if y changes twice as fast as u and u changes three times as fast as x, then we expect y to change six times as fast as x.
Ex.46 Let us try this again on another function.
y = 9x^{4} +6x^{2} +1 = (3x^{2} +1)^{2}
is the composite y = u^{2} and u = 3x^{2} + 1. Calculating derivatives. We see that
= 2 (3x^{2} + 1). 6x = 36x^{3} + 12 x
and = 36 x^{3} + 12 x
Once again,
The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x.
Ex.47 Find the derivation of
Sol. Here y = f(g(x)), where f(u) = and u = g(x) = x^{2} + 1. Since the derivatives of f and g are
f' (u) = and g'(x) = 2x,
the Chain Rule gives
= f' (g(x)).g'(x) = .g'(x) = . (2x) =
Ex.48
Ex. 49 u = 1  x^{2} and n = 1/4
(Function defined) on [1, 1]
Rule No. 12 Power Chain Rule
* If
Ex.50 = = 1 (3x 2)^{2}
= 1 (3x 2)^{2} (3) = 
In part (d) we could also have found the derivation with the Quotient Rule.
Ex.51 (a)
Sol. Here u = Ax B,
(b)
(c)log(Ax B) = .A
(d)tan (Ax + B) = sec^{2} (Ax + B).A
(e)
Note : These results are important
19. DOUBLE DIFFERENTIATION
If f is differentiable function, then its derivative f' is also a function, so f' may have a derivative of its own, denoted by . This new function f'' is called the second derivative of because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y = f(x) as
Another notation is f''(x) = D_{2} f(x).
Ex.52 If(x) = x cos x, find f'' (x)
Sol. Using the Product Rule, we have f'(x)
To find f" (x) we differentiate f'(x)
=  x cos x  sinx  sinx =  x cos x  2 sin x
20. Application of derivative Differentiation as a rate of change
is rate of change of 'y' with respect to 'x' :
For examples :
(i) v = this means velocity 'v' is rate of change of displacement 'x' with respect to time 't'
(ii) a = this means acceleration 'a' is rate of change of velocity 'v' with respect to time 't'.
(iii) this means force 'F' is rate of change of momentum 'p' with respect to time 't'.
(iv) = this means torque 't' is rate of change of angular momentum 'L' with respect to time 't'
(v) Power = this means power 'P' is rate of change of work 'W' with respect to time 't'
Ex.53 The area A of a circle is related to its diameter by the equation .
How fast is the area changing with respect to the diameter when the diameter is 10 m ?
Sol. The (instantaneous) rate of change of the area with respect to the diameter is
When D =10m, the area is changing at rate (π/2) = 5π m^{2}/m. This means that a small change ΔD m in the diameter would result in a change of about 5p ΔD m^{2} in the area of the circle.
Physical Example :
Ex.54 Boyle's Law states that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant : PV = C. Find the rate of change of volume with respect to pressure.
Sol.
Ex.55 (a) Find the average rate of change of the area of a circle with respect to its radius r as r changed from
(i) 2 to 3 (ii) 2 to 2.5 (iii) 2 to 2.1
(b) Find the instantaneous rate of change when r = 2.
(c) Show that there rate of change of the area of a circle with respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically when this is true by drawing a circle whose radius is increased by an amount Δr. How can you approximate the resulting change in area ΔA if Δr is small ?
Sol. (a) (i) 5π (ii) 4.5 π (iii) 4.1 π
(b) 4π
(c) ΔA ≈ 2 πrΔr
21. MAXIMA & MINIMA
Suppose a quantity y depends on another quantity x in a manner shown in figure. It becomes maximum at x_{1} and minimum at x_{2}. At these points the tangent to the curve is parallel to the xaxis and hence its slope is tanθ = 0. Thus, at a maximum or a minima slope
⇒
Maxima
Just before the maximum the slope is positive, at the maximum it is zero and just after the maximum it is negative. Thus, decrease at a maximum and hence the rate of change of is negative at a maximum i.e., at maximum. The quantity is the rate of change of the slope. It is written as . Conditions for maxima are : (a) (b)
Minima
Similarly, at a minimum the slope changes from negative to positive, Hence with the increases of x. The slope is increasing that means the rate of change of slope with respect to x is positive.
Hence
Conditions for minima are :
(a) (b)
Quite often it is known from the physical situation whether the quantity is a maximum or a minimum. The test on may then be omitted.
Ex.56 Find maximum or minimum values of the functions :
(A) y = 25x^{2} + 5 10x (B) y = 9 (x 3)^{2}
Sol. (A) For maximum and minimum value, we can put
or x =
Further,
or has positive value at x = . Therefore, y has minimum value at x = . Therefore, y has minimum value at x = . Substituting x = in given equation, we get
y_{min} =
(B) y = 9 (x 3)^{2} = 9 x^{2} +9 6x
or y = 6x x^{2}
For minimum or maximum value of y we will substitute
or 6 2x = 0
x = 3
To check whether value of y is maximum or minimum at x = 3 we will have to check whether is positive or negative.
or is negative at x = 3. Hence, value of y is maximum. This maximum value of y is,
y_{max} = 9 (3 3)^{2} = 9
22. INTEGRATION
Definitions :
A function F(x) is a antiderivative of a function f(x) if
F'(x) = f(x)
for all x in the domain of f. The set of all antiderivatives of f is the indefinite integral of f with respect to x, denoted by
The symbol is an integral sign. The function f is the integrand of the integral and x is the variable of integration.
For example f(x) = x^{3} then f'(x) = 3x^{2}
So the integral of 3x^{2} is x^{3}
Similarly if f(x) = x^{3} + 4
there for the general integral of 3x^{2} is x^{3} + c where c is a constant
One antiderivative F of a function f, the other antiderivatives of f differ from F by a constant. We indicate this in integral notation in the following way :
.....(i)
The constant C is the constant of integration or arbitrary constant, Equation (1) is read, "The indefinite integral of f with respect to x is F(x) + C." When we find F(x) + C, we say that we have integrated f and evaluated the integral.
Ex.57 Evaluate
Sol.
The formula x^{2} + C generates all the antiderivatives of the function 2x. The function x^{2}+ 1, x^{2} π, and
x^{2+} are all antiderivatives of the function 2x, as you can check by differentiation.
Many of the indefinite integrals needed in scientific work are found by reversing derivative formulas.
Integral Formulas
Indefinite Integral  Reversed derivated formula 
1. ,n ¹ 1, n rational (special case)  = x^{n} 
2.  
3.  
4.  
5.  (cot x) = cosec^{2} x 
6.  = sec x tan x 
7. = cosec x +C  (cosec x) = cosec x cot x 
Ex.58 Examples based on above formulas :
(a)
(b) Formula 1 with n = 5
(c) Formula 1 with n =
(d) Formula 2 with k = 2
(e) = = Formula 3 with k =
Ex.59 Right : = x sin x + cos x C
Reason : The derivative of the righthand side is the integrand :
Check : = x cos x + sin x sin x + 0 = x cos x.
Wrong : = x sin x +C
Reason : The derivative of the righthand side is not the integrand :
Check : = x cos x + sin x + 0 x cos x
Rule No. 1 Constant Multiple Rule
Ex.60 = =
Rule No.2 Sum And Difference Rule
Ex.61 Termbyterm integration
Evaluate :
Sol. If we recognize that (x^{3}/3) x^{2} +5x is an anti derivative of x^{2} 2x +5, we can evaluate the integral as
If we do not recognize the anti derivative right away, we can generate it term by term with the sum and difference Rule :
This formula is more complicated than it needs to be. If we combine C_{1}, C_{2} and C_{3} into a single constant
C = C_{1} + C_{2} + C_{3}, the formula simplifies to
and still gives all the anti derivatives there are. For this reason we recommend that you go right to the final form even if you elect to integrate term by term. Write
Find the simplest anti derivative you can for each part add the constant at the end.
Ex.62 We can sometimes use trigonometric identities to transform integrals we do not know how to evaluate into integrals. The integral formulas for sin^{2} x and cos^{2} x arise frequently in applications.
(a) =
=
(b) =
As in part (a), but with a sign change
23. Some Indefinite integrals (An arbitrary constant should be added to each of these integrals.
(a) (provided n ¹ 1) C
(b)
(c)
(d)
(e)
(f)
Ex.63 (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
24. DEFINITE INTEGRATION OR INTEGRATION WITH LIMITS
Ex.64 = 3 [4 (1)] = (3) (5) = 15
= + cos (0) = 0 + 1 = 1
Ex.65 (1)
(2)
(3)
25. APPLICATION OF DEFINITE INTEGRAL
Calculation Of Area Of A Curve.
From graph shown in figure if we divide whole area in infinitely small strips of dx width.
We take a strip at x position of dx width.
Small area of this strip dA = f(x) dx
So, the total area between the curve and xaxis = sum of area of all strips =
Let f(x) > 0 be continuous on [a,b]. The area of the region between the graph of f and the xaxis is
Ex.66 Using an area to evaluate a definite integral
Evaluate 0 < a < b.
Sol. We sketch the region under the curve y = x, a £ x £ b (figure) and see that it is a trapezoid with height (b a) and bases a and b.
The value of the integral is the area of this trapezoid :
Thus =
and so on.
Notice that x^{2}/2 is an antiderivative of x, further evidence of a connection between antiderivatives and summation.
(i) To find impulse
so implies =
Ex.67 If F = kt then find impulse at t = 3 sec.
so impulse will be area under f  t curve
=
⇒
2. To calculate work done by force :
So area under f  x curve will give the value of work done.
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Resolution of Vectors into Components Video  09:22 min 
Test: Resolution of Vectors (NCERT) Test  5 ques 
Relative Velocity in Two Dimensions Doc  2 pages 

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