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NCERT Solutions for Class 12 Maths Chapter 7 Integrals are discussed here. This chapter deals with definite and indefinite integrals. Integration class 12 also includes elementary properties of integration including basic techniques of integration. NCERT Class 12 maths chapter 7 solutions will be very helpful when you are solving the questions from NCERT books for Class 12 Maths. These NCERT Class 12 Maths solutions chapter 7 are prepared by subject matter experts that are very easy to understand. students can practice integrals class 12 ncert solutions to get good hold on the concepts.
NCERT solutions for class 12 math chapter 7 integrals are important for board exams as well as for competitive examinations like JEE Main, VITEEE, BITSAT, etc but without command, on the concepts of integrals ncert solutions meritorious marks cant be scoured. Therefore chapter 7 maths class 12 is recommended to students. Also, you can check the NCERT solutions for other Classes here.
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Also read:
>> Integration as Inverse of Differentiation: Integration is the inverse process of differentiation.
In differential calculus, we find the derivative of a given function, while in integral calculus, we find a function whose derivative is given.
Indefinite Integrals:
∫f(x) dx = F(x) + C
These integrals are called indefinite integrals or general integrals.
C is an arbitrary constant that leads to different anti-derivatives of the given function.
Multiple Anti-Derivatives:
A derivative of a function is unique, but a function can have infinite anti-derivatives or integrals.
Properties of Indefinite Integral:
∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
For any real number k, ∫k f(x) dx = k∫f(x)dx.
In general, if f1, f2, …, fn are functions and k1, k2, …, kn are real numbers, then ∫[k1f1(x) + k2f2(x) + … + knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫f2(x) dx + … + kn ∫fn(x) dx
First Fundamental Theorem of Integral Calculus:
Define the area function A(x) = ∫[a, x]f(t)dt for x ≥ a, where f is continuous on [a, b].
Then A'(x) = f(x) for every x ∈ [a, b].
Second Fundamental Theorem of Integral Calculus:
If f is a continuous function on [a, b], then ∫[a, b]f(x)dx = F(b) - F(a), where F(x) is an antiderivative of f(x).
Standard Integral Formulas:
∫xn dx = xn+1/(n+1) + C (n ≠ -1)
∫cos x dx = sin x + C
∫sin x dx = -cos x + C
∫sec2 x dx = tan x + C
∫cosec2 x dx = -cot x + C
∫sec x tan x dx = sec x + C
∫cosec x cot x dx = -cosec x + C
∫ex dx = ex + C
∫ax dx = (ax)/ln(a) + C
∫(1/x) dx = ln|x| + C
Other Integral Formulas:
∫tan x dx = ln|sec x| + C
∫cot x dx = ln|sin x| + C
∫sec x dx = ln|sec x + tan x| + C
∫cosec x dx = ln|cosec x - cot x| + C
Free download Class 12 Maths Chapter 7 Question Answer for CBSE Exam.
Class 12 Integrals NCERT solutions Exercise: 7.1
Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection.
GIven ;
So, the anti derivative of is a function of x whose derivative is .
Therefore, we have
Or, antiderivative of is .
GIven ;
So, the antiderivative of is a function of x whose derivative is .
Therefore, we have the anti derivative of is .
GIven ;
So, the anti derivative of is a function of x whose derivative is .
Therefore, we have the anti derivative of is .
GIven ;
So, the anti derivative of is a function of x whose derivative is .
Therefore, we have the anti derivative of is .
GIven ;
So, the anti derivative of is a function of x whose derivative is .
Therefore, we have the anti derivative of is .
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given integral ;
or
, where C is any constant value.
Given intergral ;
or
Or, , where C is any constant value.
Given intergral ;
or
Or, , where C is any constant value.
Given integral
It can be written as
Taking common out
Now, cancelling out the term from both numerator and denominator.
Splitting the terms inside the brackets
Given intergral ;
or
, where C is any constant value.
Given intergral ;
or
, where C is any constant value.
Given integral ;
splitting the integral as the sum of three integrals
, where C is any constant value.
Given integral ;
Using the integral of trigonometric functions
, where C is any constant value.
Given integral ;
, where C is any constant value.
Given integral ;
Using antiderivative of trigonometric functions
, where C is any constant value.
Given to find the anti derivative or integral of ;
, where C is any constant value.
Hence the correct option is (C).
If such that f (2) = 0. Then f (x) is
Given that the anti derivative of
So,
Now, to find the constant C;
we will put the condition given, f (2) = 0
or
Therefore the correct answer is A .
Class 12 Integrals NCERT solutions Exercise: 7.2
Given to integrate function,
Let us assume
we get,
now back substituting the value of
as is positive we can write
Given to integrate function,
Let us assume
we get,
Given to integrate function,
Let us assume
we get,
Back substituting the value of t we get,
Given to integrate function,
Let us assume
we get,
Now, by back substituting the value of t,
Given to integrate function,
Let us assume
we get,
Now, by back substituting the value of t,
Given function ,
Assume the 19634
Back substituting the value of t in the above equation.
or, , where C is any constant value.
Given function ,
Assume the
Or
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t in the above equation,
, where C is any constant value.
Given function ,
Can be written in the form:
Assume the
, where C is any constant value.
Given function ,
Assume the so,
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
Now back substituting the value of t ;
, where C is any constant value.
Given function ,
Assume the
Now back substituting the value of t ;
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Simplifying it by dividing both numerator and denominator by , we obtain
Assume the
Now, back substituting the value of t,
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t,
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t,
or , where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
or simplified as
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
or simplified as
Assume the
Now, back substituted the value of t.
where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function
Assume that
Now solving the assumed integral;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now solving the assumed integral;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now solving the assumed integral;
Multiplying numerator and denominator by ;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now, back substituting the value of t,
Given function
Simplifying to solve easier;
Assume that
Now, back substituting the value of t,
Given function
Assume that
......................(1)
Now to solve further we take
So, from the equation (1), we will get
Now back substitute the value of u,
and then back substituting the value of t,
Given integral
Taking the denominator
Now differentiating both sides we get
Back substituting the value of t,
Therefore the correct answer is D.
Class 12 Integrals NCERT Solutions Exercise: 7.3
using the trigonometric identity
we can write the given question as
=
Using identity
, therefore the given integral can be written as
Using identity
Again use the same identity mentioned in the first line
The integral can be written as
Let
Now, replace the value of t, we get;
rewrite the integral as follows
Let
......(replace the value of t as )
Using the formula
we can write the integral as follows
Using identity
we can write the following integral as
=
We know the identities
Using the above relations we can write
The integral is rewritten using trigonometric identities
can be written as follows using trigonometric identities
Therefore,
now using the identity
now using the below two identities
the value
.
the integral of the given function can be written as
Using trigonometric identities we can write the given integral as follows.
We know that,
Using this identity we can rewrite the given integral as
Therefore integration of =
.....................(i)
Let assume
So, that
Now, the equation (i) becomes,
the given question can be rearranged using trigonometric identities
Therefore, the integration of = ...................(i)
Considering only
let
now the final solution is,
now splitting the terms we can write
Therefore, the integration of
The integral of the above equation is
Thus after evaluation, the value of integral is tanx+ c
Let
We can write 1 =
Then, the equation can be written as
put the value of tan = t
So, that
we know that
therefore,
let
Now the given integral can be written as
using the trigonometric identities we can evaluate the following integral as follows
Using the trigonometric identities following integrals can be simplified as follows
The correct option is (A)
On reducing the above integral becomes
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.4
The given integral can be calculated as follows
Let
, therefore,
let suppose 2x = t
therefore 2dx = dt
.................using formula
Let assume 5x =t,
then 5dx = dt
The above result is obtained using the identity
let
then
using the special identities we can simplify the integral as follows
We can write above eq as
............................................(i)
for let
Now, by using eq (i)
The integration can be down as follows
let
........................using
the above equation can be also written as,
let 1+x = t
then dx = dt
therefore,
the denominator can be also written as,
therefore
Let x+3 = t
then dx =dt
......................................using formula
(x-1)(x-2) can be also written as
=
=
therefore
let suppose
Now,
.............by using formula
let
By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0
Let
let
By comparing the coefficients and constant term on both sides, we get;
A=1/2 and B=2
then
let
By comparing the coefficients and constants we get the value of A and B
A = and B =
NOW,
...........................(i)
put
Thus
let
By comparing the coefficients and constants on both sides, we get
A =3 and B =34
....................................(i)
Considering
let
Now consider
here the denominator can be also written as
Dr =
Now put the values of and in eq (i)
let
By equating the coefficients and constant term on both sides we get
A = -1/2 and B = 4
(x+2) = -1/2(4-2x)+4
....................(i)
Considering
let
now,
put the value of and
...........(i)
take
let
considering
putting the values in equation (i)
Let
By comparing the coefficients and constant term, we get;
A = 1/2 and B =4
..............(i)
put
let
On comparing, we get
A =5/2 and B = -7
...........................................(i)
put
The correct option is (B)
the denominator can be written as
now,
The following integration can be done as
The correct option is (B)
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.5
Question:1 Integrate the rational functions
Answer:
Given function
Partial function of this function:
Now, equating the coefficients of x and constant term, we obtain
On solving, we get
Given function
The partial function of this function:
Now, equating the coefficients of x and constant term, we obtain
On solving, we get
Given function
Partial function of this function:
.(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Partial function of this function:
.....(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Partial function of this function:
...........(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Integral is not a proper fraction so,
Therefore, on dividing by , we get
Partial function of this function:
...........(1)
Now, substituting respectively in equation (1), we get
No, substituting in equation (1) we get
Given function
Partial function of this function:
Now, equating the coefficients of and the constant term, we get
and
On solving these equations, we get
From equation (1), we get
Now, consider ,
and we will assume
So,
or
Given function
Partial function of this function:
Now, putting in the above equation, we get
By equating the coefficients of and constant term, we get
then after solving, we get
Therefore,
Given function
can be rewritten as
Partial function of this function:
................(1)
Now, putting in the above equation, we get
By equating the coefficients of and , we get
then after solving, we get
Therefore,
Given function
can be rewritten as
The partial function of this function:
Equating the coefficients of , we get
Therefore,
Given function
can be rewritten as
The partial function of this function:
Now, substituting the value of respectively in the equation above, we get
Therefore,
Given function
As the given integral is not a proper fraction.
So, we divide by , we get
can be rewritten as
....................(1)
Now, substituting in equation (1), we get
Therefore,
Given function
can be rewritten as
....................(1)
Now, equating the coefficient of and constant term, we get
, , and
Solving these equations, we get
Therefore,
Given function
can be rewritten as
Now, equating the coefficient of and constant term, we get
and ,
Solving these equations, we get
Therefore,
Given function
can be rewritten as
The partial fraction of above equation,
Now, equating the coefficient of and constant term, we get
and
and
Solving these equations, we get
Therefore,
[Hint: multiply numerator and denominator by and put ]
Given function
Applying Hint multiplying numerator and denominator by and putting
Putting
can be rewritten as
Partial fraction of above equation,
................(1)
Now, substituting in equation (1), we get
Given function
Applying the given hint: putting
We get,
Partial fraction of above equation,
................(1)
Now, substituting in equation (1), we get
Back substituting the value of t in the above equation, we get
Given function
We can rewrite it as:
Partial fraction of above equation,
Now, equating the coefficients of and constant term, we get
, , ,
After solving these equations, we get
Given function
Taking
The partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Given function
So, we multiply numerator and denominator by , to obtain
Now, putting
we get,
Taking
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Back substituting the value of t,
Given function
So, applying the hint: Putting
Then
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Now, back substituting the value of t,
Given integral
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Therefore, the correct answer is B.
Given integral
Partial fraction of above equation,
Now, equating the coefficients of and the constant term, we get
, ,
We have the values,
Therefore, the correct answer is A.
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.6
Question:1 Integrate the functions
Answer:
Given function is
We will use integrate by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Again use integration by parts in
Put this value in our equation
we will get,
Therefore, answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Now, we need to integrate
Put this value in our equation
Therefore, the answer is
Given function is
We will use integration by parts method
Put this value in our equation
Therefore, the answer is
Given function is
We will use integration by parts method
Now, we need to integrate
Put this value in our equation
Therefore, the answer is
Given function is
we will use integration by parts method
Therefore, answer is
Consider
So, we have then:
After taking as a first function and as second function and integrating by parts, we get
Or,
Consider
So, we have then:
After taking as a first function and as second function and integrating by parts, we get
Consider
So, we have then:
After taking as a first function and as second function and integrating by parts, we get
Consider
So, we have then:
After taking as a first function and as second function and integrating by parts, we get
Consider
So, we have then:
Let us take ....................(1)
Where, and
So,
After taking as a first function and as second function and integrating by parts, we get
....................(2)
After taking as a first function and as second function and integrating by parts, we get
................(3)
Now, using the two equations (2) and (3) in (1) we get,
Let suppose
we know that,
Thus, the solution of the given integral is given by
Let suppose
by rearranging the equation, we get
let
It is known that
therefore the solution of the given integral is
Let
substitute and
let
It is known that
Therefore the solution of the given integral is
It is known that
let
Therefore the required solution of the given above integral is
It is known that
So, By adjusting the given equation, we get
to let
Therefore the required solution of the given integral is
let
Taking as a first function and as a second function, by using by parts method
we know that,
from above integral
let
thus, the solution of the above integral is
NCERT class 12 maths ch 7 question answer Exercise: 7.7
Given function ,
So, let us consider the function to be;
Then it is known that,
Therefore,
Given function to integrate
Now we can rewrite as
As we know the integration of this form is
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
a
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
(A)
(B)
(C)
(D)
As we know that,
So,
Therefore the correct answer is A.
(A)
(B)
(C)
(D)
Given integral
So, let us consider the function to be;
And we know that,
Therefore the correct answer is D.
NCERT class 12 maths ch 7 question answer - Exercise:7.8
We know that,
This is how the integral is evaluated using limit of a sum
We know that
let
Here a = 0, b = 5 and
therefore
We know that
here a = 2 and b = 3 , so h = 1/n
Let
for the second part, we already know the general solution of
So, here a = 1 and b = 4
therefore
So,
let
We know that
Here a =-1, b = 1 and
therefore h = 2/n
By using sum of n terms of GP ....where a = 1st term and r = ratio
.........using
It is known that,
..........................( )
NCERT class 12 maths ch 7 question answer - Exercise:7.9
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
Multiplying by 5 both in numerator and denominator:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
Putting which gives,
As, and as .
So, we have now:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
So, we can rewrite the integral as;
where . ................(1)
Now, consider
Take numerator
We now equate the coefficients of x and constant term, we get
Now take denominator
Then we have
Then substituting the value of in equation (1), we get
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
or we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given definite integral
Consider
we have then the function of x, as
By applying the second fundamental theorem of calculus, we will get
Therefore the correct answer is D.
(A)
(B)
(C)
(D)
Given definite integral
Consider
Now, putting
we get,
Therefore we have,
we have the function of x , as
So, by applying the second fundamental theorem of calculus, we get
Therefore the correct answer is C.
NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.10
let
when x = 0 then t = 1 and when x =1 then t = 2
let
when and
using the above substitution we can evaluate the integral as
let
when x = 0 then and when x = 1 then
Taking as a first function and as a second function, by using by parts method
Let
when x = 0 then t = and when x=2 then t = 2
let
when x=0 then t = 1 and when x= then t = 0
By adjusting, the denominator can also be written as
Now,
let
when x= 0 then t =-1/2 and when x =2 then t = 3/2
On rationalisation, we get
the Dr can be written as
and put x+1 = t then dx =dt
when x= -1 then t = 0 and when x = 1 then t = 2
let
when x = 1 then t = 2 and when x = 2 then t= 4
let
(A) 6
(B) 0
(C) 3
(D) 4
The value of integral is (A) = 6
let
now, when x = 1/3, t = 8 and when x = 1 , t = 0
therefore
(A)
(B)
(C)
(D)
The correct answer is (B) =
by using by parts method,
So,
NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.11
We have ............................................................. (i)
By using
We get :-
or
................................................................ (ii)
Adding both (i) and (ii), we get :-
or
or
or
or
We have .......................................................................... (i)
By using ,
We get,
or .......................................................(ii)
Adding (i) and (ii), we get,
or
or
Thus
We have ..................................................................(i)
By using :
We get,
or . ............................................................(ii)
Adding (i) and (ii), we get :
or
or
Thus
We have ..................................................................(i)
By using :
We get,
or . ............................................................(ii)
Adding (i) and (ii), we get :
or
or
Thus
We have,
For opening the modulas we need to define the bracket :
If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).
So the integral becomes :-
or
This gives
We have,
For opening the modulas we need to define the bracket :
If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).
So the integral becomes:-
or
This gives
We have
U sing the property : -
We get : -
or
or
or
or
or
We have
By using the identity
We get,
or
or
or
or
or
or
We have
By using the identity
We get :
or
or
or
or
or
We have
or
or ..............................................................(i)
By using the identity :
We get :
or ....................................................................(ii)
Adding (i) and (ii) we get :-
or
or
We have
We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.
Also,
So,
or
or
We have ..........................................................................(i)
By using the identity :-
We get,
or ............................................................................(ii)
Adding both (i) and (ii) we get,
or
or
or
We have
We know that is an odd function.
So the following property holds here:-
Hence
We have
I t is known that :-
If f (2a - x) = f(x)
If f (2a - x) = - f(x)
Now, using the above property
Therefore,
We have ................................................................(i)
By using the property :-
We get ,
or ......................................................................(ii)
Adding both (i) and (ii), we get
Thus I = 0
We have .....................................................................................(i)
By using the property:-
We get,
or
....................................................................(ii)
Adding both (i) and (ii) we get,
or
or
or ........................................................................(iii)
or ........................................................................(iv)
or .....................................................................(v)
Adding (iv) and (v) we get,
We have ................................................................................(i)
By using, we get
We get,
.................................................................(ii)
Adding (i) and (ii) we get :
or
or
We have,
For opening the modulas we need to define the bracket :
If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).
So the integral becomes:-
or
This gives
Let ........................................................(i)
This can also be written as :
or ................................................................(ii)
Adding (i) and (ii), we get,
or
(A) 0
(B) 2
(C)
(D) 1
We have
This can be written as :
Also if a function is even function then
And if the function is an odd function then :
Using the above property I become:-
or
or
We have
.................................................................................(i)
By using :
We get,
or .............................................................................(ii)
Adding (i) and (ii), we get:
or
Thus
NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise
Firstly we will simplify the given equation :-
Let
By solving the equation and equating the coefficients of x 2 , x and the constant term, we get
Thus the integral can be written as :
or
At first we will simplify the given expression,
or
Now taking its integral we get,
or
or
Let
Using the above substitution we can write the integral is
or
or
or
or
For the simplifying the expression, we will multiply and dividing it by x -3 .
We then have,
Now, let
Thus,
or
[Hint: , put ]
Put
We get,
or
or
or
Now put in the above result :
Let us assume that :
Solving the equation and comparing coefficients of x 2 , x and the constant term.
We get,
Thus the equation becomes :
or
or
or
or
We have,
Assume :-
Putting this in above integral :
or
or
or
or
We have the given integral
Assume
So, this substitution gives,
or
We have
Simplifying the given expression, we get :
or
or
or
Thus,
and
For simplifying the given equation, we need to multiply and divide the expression by .
Thus we obtain :
or
or
or
Thus integral becomes :
or
or
Given that to integrate
Let
the required solution is
we have to integrate the following function
Let
using this we can write the integral as
Given,
Let
Now, Using partial differentiation,
Equating the coefficients of and constant value,
A + C = 0 C = -A
B + D = 0 B = -D
4A + C =0 4A = -C 4A = A A = 0 = C
4B + D = 1 4B – B = 1 B = 1/3 = -D
Putting these values in equation, we have
Given,
(let)
Let
using the above substitution the integral is written as
Given the function to be integrated as
Let
Let
Given,
Let
Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt
Now we can write the ntegral as
Given,
Let
We know the identity that
sin (A+B) = sin A cos B + cos A sin B
We have
or
or
or
or
or
Thus
Now we will solve I'.
Put x = t 2 .
Differentiating the equation wrt x, we get
Thus
or
Using integration by parts, we get :
or
We know that
Thus it becomes :
So I come to be :-
Given,
= I (let)
Let
using the above substitution we can write the integral as
Given to evaluate
now the integral becomes
Let tan x = f(x)
Given,
using partial fraction we can simplify the integral as
Let
Equating the coefficients of x, x 2 and constant value, we get:
A + C = 1
3A + B + 2C = 1
2A+2B+C =1
Solving these:
A= -2, B=1 and C=3
We have
Let us assume that :
Differentiating wrt x,
Substituting this in the original equation, we get
or
or
Using integration by parts , we get
or
or
Putting all the assumed values back in the expression,
or
Here let's first reduce the log function.
Now, let
So our function in terms if new variable t is :
now let's solve this By using integration by parts
Since, we have multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,
So,
Here let's use the property
so,
First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)
Let' divide both numerator and denominator by
Now lets change the variable
the limits will also change since the variable is changing
So, the integration becomes:
Lets first simplify the function.
As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,
AS we can write square of sec in term of tan,
Now let's calculate the integral of the second function, (we already have calculated the first function)
let
here we are changing the variable so we have to calculate the limits of the new variable
when x = 0, t = 2tanx = 2tan(0)=0
when
our function in terms of t is
Hence our total solution of the function is
Here first let convert sin2x as the angle of x ( sinx, and cosx)
Now let's remove the square root form function by making a perfect square inside the square root
Now let
,
since we are changing the variable, limit of integration will change
our function in terms of t :
First, let's get rid of the square roots from the denominator,
First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt
So,
Now since we are changing the variable, the new limit of the integration will be,
when x = 0, t = cos0-sin0=1-0=1
when
Now,
Hence our function in terms of t becomes,
Let I =
Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so
Now the important step here is to change the limit of the integration as we are changing the variable.so,
So our function becomes,
Now, let's integrate this by using integration by parts method,
Let I = -(i)
Replacing x with ( -x),
- (ii)
Adding (i) and (ii)
Given integral
So, we split it in according to intervals they are positive or negative.
Now,
as is positive in the given x -range
Therefore,
as is in the given x -range and in the range
Therefore,
as is in the given x -range and in the range
Therefore,
So, We have the sum
.
L.H.S =
We can write the numerator as [(x+1) -x]
= RHS
Hence proved.
Integrating I by parts
Applying Limits from 0 to 1
Hence proved I = 1
The Integrand g(x) therefore is an odd function and therefore
For I 2 let cosx=t, -sinxdx=dt
The limits change to 0 and 1
I 1 -I 2 =2/3
Hence proved.
The integral is written as
Hence Proved
Integrating by parts we get
For I 2 take 1-x 2 = t 2 , -xdx=tdt
Hence Proved
Question:40 Evaluate as a limit of a sum.
As we know
where b-a=hn
In the given problem b=1, a=0 and
(A)
(B)
(C)
(D)
the above integral can be re arranged as
let e x =t, e x dx=dt
(A) is correct
(A)
(B)
(C)
(D)
cos2x=cos 2 x-sin 2 x
let sinx+cosx=t,(cosx-sinx)dx=dt
hence the given integral can be written as
B is correct
(A)
(B)
(C)
(D)
As we know
Using the above property we can write the integral as
Answer (D) is correct
(A) 1
(B) 0
(C) -1
(D)
as
Now the integral can be written as
(B) is correct.
If you are looking for integrals class 12 NCERT solutions of exercises then they are listed below.
Integrals has 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own. If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals. In this chapter, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercise. Here, the NCERT solutions for class 12 maths chapter 7 integrals are solved and explained in detail to develop a grip on the topic. Here, you will learn two types of integrals: Definite integral and Indefinite integral and also learn their properties and formulas.
Definite Integral | Indefinite Integral | |
Definition | A definite Integral has upper and lower limits if 'a' and 'b' are the limits or boundaries. The definite integral of f(x) is a number, not function . | An integral without upper limit and lower limit. It is also an antiderivative. The indefinite integral of f(x) is a function not number. |
Expression |
|
7.1 Introduction
7.2 Integration as an Inverse Process of Differentiation
7.2.1 Geometrical interpretation of indefinite integral
7.2.2 Some properties of indefinite integral
7.2.3 Comparison between differentiation and integration
7.3 Methods of Integration
7.3.1 Integration by substitution
7.3.2 Integration using trigonometric identities
7.4 Integrals of Some Particular Functions
7.5 Integration by Partial Fractions
7.6 Integration by Parts
7.7 Definite Integral
7.7.1 Definite integral as the limit of a sum
7.8 Fundamental Theorem of Calculus
7.8.1 Area function
7.8.2 First fundamental theorem of integral calculus
7.8.3 Second fundamental theorem of integral calculus
7.9 Evaluation of Definite Integrals by Substitution
7.10 Some Properties of Definite Integral
Chapter 1 | Relations and Functions |
Chapter 2 | Inverse Trigonometric Functions |
Chapter 3 | Matrices |
Chapter 4 | Determinants |
Chapter 5 | Continuity and Differentiability |
Chapter 6 | Application of Derivatives |
Chapter 7 | Integrals |
Chapter 8 | Application of Integrals |
Chapter 9 | Differential Equations |
Chapter 10 | Vector Algebra |
Chapter 11 | Three Dimensional Geometry |
Chapter 12 | Linear Programming |
Chapter 13 | Probability |
The NCERT chapter 7 class 12 maths offers several key features to aid students in their understanding and mastery of this topic. Some of these features include
Detailed explanation: The chapter 7 class 12th maths solutions provide a comprehensive and in-depth explanation of the concepts of integrals, making it easier for students to grasp the subject.
Step-by-Step Solutions: The integration class 12 ncert solutions break down complex problems into simpler steps, making it easier for students to follow and understand the process.
Clear Diagrams: The integrals class 12 solutions make use of clear and informative diagrams to help students visualize the concepts, making it easier for them to comprehend the material.
Plenty of Practice Problems: The ch 7 maths class 12 include a large number of practice problems to help students strengthen their knowledge and skills.
Accurate answers: The class 12 ch 7 maths ncert solutions are checked and verified by experts to ensure accuracy, helping students avoid mistakes and improve their grades.
Also read,
The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.
Integrals have 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own.
If you are not able to, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals.
In this NCERT Class 12 Maths solutions chapter 7, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercises. Here, the Class 12 Maths Chapter 7 NCERT solutions are solved and explained in detail to develop a grip on the topic.
You are expected to remember all formulas of differentiation, and then can start with basic integration
Try to relate differentiation formulas with integrations formulas because it will help you to remember all the integration formulas
When you are done with basic integration, you should learn different methods of integration for different types of functions. If you find difficulties in learning the method you should learn one method at a time to solve a particular type of problem
After learning the different methods of integration, you should be able to determine which method will be used to solve a particular type of problem
When you become good with indefinite integrals, move to the definite integrals and learn some properties to solve definite integrals. NCERT Class 12 Maths solutions chapter 7 integrals will help you for the same
This chapter requires a lot of practice. First, solve all the NCERT textbook questions, then, you can take the help of NCERT solutions for class 12 maths chapter 7 integrals.
If you have solved all NCERT questions, you can solve CBSE previous year papers also to get familiar with the type of questions which are asked in previous years.
NCERT solutions class 12 maths chapter 12 pdf download will be made available soon. Till then you can save the webpage and practice these solutions offline.
Students consider Integration which is integrals and applications of integration are the most difficult chapter in CBSE class 12 maths but with the regular practice of NCERT problems you will be able to have a strong grip on this chapter also. it's true that there is no substitute for hard work but the right strategy and quality study material are also essential to get command of this chapter, therefore, NCERT exercises are recommended for practice.
NCERT solutions are created by the expert team of careers360 who know how best to write answers in the board exam in order to get good marks. Integrating their techniques can benefit in obtaining meritorious marks. Sometimes students do not understand where s/he are making a mistake, NCERT solutions can help them to understand that. the practice of a lot of questions and their solution make you confident and help you in getting an in-depth understanding of concepts. Therefore NCERT solutions are very helpful for students.
Integrals have 13 % weightage in CBSE class 12th board final examination. having 13% weightage, Integral become very students for CBSE aspirant but it demand lot of practice. Interested students can refer to integrals class 12 solutions.
The NCERT Solutions for maths ch 7 maths class 12 is not complicated to understand. It is a fascinating topic in Class 12 that is also relevant at higher education levels. A solid understanding of integral formulas will enable students to solve integration problems effectively. A thorough knowledge of derivatives is crucial for comprehending the concepts of integral calculus with ease. For ease, students can study integrals class 12 ncert solutions pdf both online and offline.
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
Hello dear,
If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.
As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.
Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.
Believe in Yourself! You can make anything happen
All the very best.
Hello Student,
I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.
You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.
All the best.
If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.
Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.
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Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.
For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.
In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.
Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.
Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.
Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).
Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian.
The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.
Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.
A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.
A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.
A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment.
An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.
An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.
Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack
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