Chapter 7 Integrals

8/12/2019 Chapter 7 Integrals

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Class XII Chapter 7 Integrals Maths

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Exercise 7.1

Question 1:

sin 2x

Answer

The anti derivative of sin 2xis a function ofxwhose derivative is sin 2x.It is known that,

Therefore, the anti derivative of

Question 2:

Cos 3x

Answer

The anti derivative of cos 3xis a function ofxwhose derivative is cos 3x.

It is known that,

Therefore, the anti derivative of .

Question 3:

e2x

Answer

The anti derivative of e2x is the function ofxwhose derivative is e2x.

It is known that,


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Question 4:

Answer

The anti derivative of is the function ofx whose derivative is .

It is known that,


Question 5:

Answer

The anti derivative of is the function ofxwhose derivative is

.It is known that,


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Therefore, the anti derivative of is .

Question 6:

Answer

Question 7:

Answer

Question 8:

Answer

Cl XII Ch t 7 I t l M th


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Question 9:

Answer

Question 10:

Answer



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Question 11:

Answer

Question 12:

Answer



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Question 13:

Answer

On dividing, we obtain

Question 14:

Answer

Question 15:

Answer



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Question 16:

Answer

Question 17:

Answer



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p g

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Question 18:

Answer

Question 19:

Answer



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g

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Question 20:

Answer

Question 21:

The anti derivative of equals

(A) (B)

(C) (D)

Answer



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Hence, the correct Answer is C.

Question 22:

If such thatf(2) = 0, then f(x) is

(A) (B)

(C) (D)

Answer

It is given that,

Anti derivative of



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Also,

Hence, the correct Answer is A.



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Exercise 7.2

Question 1:

Answer

Let = t

2x dx= dt

Question 2:

Answer

Let log |x| = t



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Question 3:

Answer

Let 1 + logx = t

Question 4:

sinxsin (cosx)

Answer



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sinxsin (cosx)

Let cosx= t

sinx dx = dt

Question 5:

Answer

Let

2adx = dt



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Question 6:

Answer

Let ax + b = t

adx = dt

Question 7:

Answer

Let



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dx = dt

Question 8:

Answer

Let 1 + 2x2= t

4xdx = dt



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Question 9:

Answer

Let

(2x+ 1)dx = dt



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Question 10:

Answer

Let

Question 11:

Answer

Let

dx = dt



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Question 12:

Answer

Let



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Question 13:

Answer

Let

9x2dx = dt



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Question 14:

Answer

Let logx = t

Question 15:

Answer


L t


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Let

8x dx= dt

Question 16:

Answer

Let

2dx = dt



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Question 17:

Answer

Let

2xdx = dt

Question 18:

Answer

Let



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Question 19:

Answer

Dividing numerator and denominator by ex, we obtain

Let


Question 20:


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Answer

Let

Question 21:

Answer

Let 2x 3 = t

2dx = dt



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Question 22:

Answer

Let 7 4x= t

4dx = dt

Question 23:

Answer

Let



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Question 24:

Answer

Let


Question 25:


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Answer

Let

Question 26:

Answer

Let



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Question 27:

Answer

Let sin 2x= t

Question 28:

Answer

Let



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cosx dx= dt

Question 29:

cotxlog sinx

Answer

Let log sinx= t

Question 30:

Answer

Let 1 + cosx = t


i d dt


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sinx dx= dt

Question 31:

Answer

Let 1 + cosx= t

sinxdx = dt

Question 32:

Answer



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Let sinx+ cosx= t(cosx sinx) dx= dt

Question 33:

Answer



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Put cosx sinx= t(sinx cosx) dx= dt

Question 34:

Answer



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Question 35:

AnswerLet 1 + logx= t


Question 36:


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Answer

Let

Question 37:

Answer

Letx4= t

4x3dx = dt



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Let

From (1), we obtain

Question 38:

equals

Answer

Let



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Hence, the correct Answer is D.

Question 39:

equals

A.

B.

C.

D.

Answer

Hence, the correct Answer is B.


Exercise 7.3


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Question 1:

Answer

Question 2:

Answer

It is known that,


Question 3:


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cos 2xcos 4xcos 6x

Answer

It is known that,

Question 4:

sin3(2x+ 1)

Answer

Let



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Question 5:

sin3xcos3x

Answer

Question 6:

sinxsin 2xsin 3x

Answer


It is known that,


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Question 7:

sin 4xsin 8x

Answer

It is known that,



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Question 8:

Answer

Question 9:

Answer



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Question 10:sin4x

Answer



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Question 11:

cos42x

Answer



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Question 12:

Answer


Question 13:


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Answer

Question 14:

Answer



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Question 15:

Answer



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Question 16:

tan4x

Answer


From equation (1), we obtain


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Question 17:

Answer

Question 18:

Answer

Question 19:


Answer


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Question 20:

Answer



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Question 21:

sin1(cosx)

Answer



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It is known that,

Substituting in equation (1), we obtain



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Question 22:

Answer


Question 23:

is equal to


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is equal to

A.tanx+ cotx+ C

B.tanx+ cosecx+ CC. tanx+ cotx+ C

D. tanx+ secx+ C

Answer


Question 24:

equals

A. cot (exx) + C

B.tan (xex) + C

C.tan (ex) + C

D. cot (ex) + C

Answer

Let exx= t



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Exercise 7.4

Question 1:


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AnswerLetx3= t

3x2dx= dt

Question 2:

Answer

Let 2x= t

2dx= dt



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Question 3:

Answer

Let 2 x = t

dx= dt

Question 4:

Answer

Let 5x=t


5dx= dt


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Question 5:

Answer

Question 6:

Answer

Letx3= t


3x2dx= dt


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Question 7:

Answer

From (1), we obtain



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Question 8:

Answer

Letx3= t

3x2

dx= dt

Question 9:

Answer

Let tanx=t

sec2xdx= dt



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Question 10:

Answer

Question 11:

Answer



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Question 12:

Answer


Question 13:


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Answer

Question 14:


Answer


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Question 15:

Answer



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Question 16:

Answer

Equating the coefficients ofxand constant term on both sides, we obtain


4A= 4 A= 1


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A+ B= 1 B= 0

Let 2x2+x 3 = t

(4x+ 1) dx = dt

Question 17:

Answer


From (1), we obtain



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Question 18:

Answer

Equating the coefficient ofxand constant term on both sides, we obtain



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Substituting equations (2) and (3) in equation (1), we obtain



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Question 19:

Answer

Equating the coefficients ofxand constant term, we obtain

2A= 6 A= 3

9A+ B= 7 B= 34

6x+ 7 = 3 (2x 9) + 34



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Substituting equations (2) and (3) in (1), we obtain

Question 20:

Answer




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Using equations (2) and (3) in (1), we obtain


Q ti 21


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Question 21:

Answer

Letx2+ 2x+3 = t

(2x+ 2) dx=dt



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Question 22:

Answer




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Substituting (2) and (3) in (1), we obtain

Question 23:

Answer




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Question 24:


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equals

A.xtan1(x+ 1) + C

B.tan 1(x+ 1) + C

C.(x+ 1) tan1x+ C

D. tan1x+ C

Answer


Question 25:

equals

A.

B.

C.


D.

Answer


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Exercise 7.5

Question 1:

Answer


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Let


A+ B = 1

2A+B = 0

On solving, we obtain

A= 1 and B= 2

Question 2:

Answer

Let



A+B = 0

3A+ 3B= 1



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Question 3:

Answer

Let

Substitutingx= 1, 2, and 3 respectively in equation (1), we obtain

A= 1, B= 5, and C= 4


Question 4:

Answer

L t


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Let

Substitutingx= 1, 2, and 3 respectively in equation (1), we obtain

Question 5:

Answer

Let

Substitutingx= 1 and 2 in equation (1), we obtain

A= 2 and B= 4



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Question 6:

Answer

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 x2

) byx(1 2x), we obtain

Let

Substitutingx= 0 and in equation (1), we obtain

A = 2 andB = 3

Substituting in equation (1), we obtain



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Question 7:

Answer

Let

Equating the coefficients ofx2,x, and constant term, we obtain

A+ C= 0

A+ B= 1

B+ C= 0

On solving these equations, we obtain




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Question 8:

Answer

Let

Substitutingx= 1, we obtain

Equating the coefficients ofx2and constant term, we obtain

A+ C= 0

2A+ 2B+ C= 0




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Question 9:

Answer

Let

Substitutingx= 1 in equation (1), we obtain

B= 4

Equating the coefficients ofx2andx, we obtain

A+ C= 0

B 2C= 3




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Question 10:

Answer

Let

Equating the coefficients ofx2andx, we obtain



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Question 11:

Answer

Let

Substitutingx = 1, 2, and 2 respectively in equation (1), we obtain

Question 12:


Answer

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3+x + 1) byx2 1, we obtain

Let


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Let

Substitutingx = 1 and 1 in equation (1), we obtain

Question 13:

Answer

Equating the coefficient ofx2,x, and constant term, we obtain

A B= 0

B C= 0

A+ C= 2



A= 1, B= 1, and C= 1


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Question 14:

Answer

Equating the coefficient ofxand constant term, we obtain

A= 3

2A+ B = 1 B= 7



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Question 15:

Answer

Equating the coefficient ofx3,x2,x, and constant term, we obtain




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Question 16:

[Hint: multiply numerator and denominator byxn 1and putxn= t]

Answer

Multiplying numerator and denominator byxn 1, we obtain

Substituting t= 0, 1 in equation (1), we obtain

A= 1 and B= 1



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Question 17:

[Hint: Put sinx= t]

Answer

Substituting t= 2 and then t= 1 in equation (1), we obtain

A= 1 and B= 1



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Question 18:

Answer

Equating the coefficients ofx3,x2,x, and constant term, we obtain

A+ C= 0

B+ D= 4

4A+ 3C= 0

4B+ 3D= 10


A= 0, B= 2, C= 0, and D= 6



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Question 19:

Answer

Letx2= t2xdx= dt

Substituting t = 3 and t = 1 in equation (1), we obtain



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Question 20:

Answer

Multiplying numerator and denominator byx3, we obtain

Letx4=t4x3dx= dt


Substitutingt = 0 and 1 in (1), we obtain

A= 1 and B= 1


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Question 21:

[Hint: Put ex= t]

Answer

Let ex= t exdx= dt


Substituting t= 1 and t= 0 in equation (1), we obtain

A= 1 and B= 1


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Question 22:

A.

B.

C.

D.

Answer

Substitutingx= 1 and 2 in (1), we obtain

A= 1 and B= 2



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Question 23:

A.

B.

C.

D.

Answer


A+ B= 0

C= 0

A= 1


A = 1, B= 1, and C= 0




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Exercise 7.6

Question 1:

xsinx

Answer

Let I=

Takingxas first function and sinxas second function and integrating by parts, we

obtain


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obtain

Question 2:

Answer

Let I=

Takingxas first function and sin 3xas second function and integrating by parts, we

obtain

Question 3:


Answer

Let

Takingx2as first function and exas second function and integrating by parts, we obtain


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Again integrating by parts, we obtain

Question 4:

xlogx

Answer

Let

Taking logxas first function andxas second function and integrating by parts, we

obtain


Question 5:

xlog 2x

Answer

Let

Taking log 2xas first function andxas second function and integrating by parts, we

obtain


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Question 6:

x2 logx

Answer

Let

Taking logxas first function andx2as second function and integrating by parts, we

obtain


Question 7:

Answer

Let

Taking as first function andxas second function and integrating by parts, we

obtain


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Question 8:

Answer

Let


Taking as first function andxas second function and integrating by parts, we

obtain


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Question 9:

Answer

Let

Taking cos1xas first function andxas second function and integrating by parts, we

obtain



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Question 10:

Answer

Let

Taking as first function and 1 as second function and integrating by parts, we

obtain


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obtain

Question 11:

Answer

Let


Taking as first function and as second function and integrating by parts,

we obtain


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Question 12:

Answer

Let

Takingxas first function and sec2xas second function and integrating by parts, we

obtain

Question 13:

Answer


Let


obtain


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Question 14:

Answer


obtain




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Question 15:

Answer

Let

Let I= I1+ I2 (1)

Where, and

Taking logxas first function andx

2

as second function and integrating by parts, weobtain

Taking logxas first function and 1 as second function and integrating by parts, we

obtain




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Question 16:

Answer

Let

Let

It is known that,

Question 17:


Answer

Let


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Let

It is known that,

Question 18:

Answer



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Let

It is known that,


Question 19:


Answer

Also, let

It is known that,


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Question 20:

Answer

Let

It is known that,

Question 21:

Answer


Let

Integrating by parts, we obtain



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g g g y p

Question 22:

Answer

Let


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Question 24:

equals


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equals

Answer

Let

Also, let

It is known that,



Exercise 7.7

Question 1:

Answer


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Question 2:

Answer


Question 3:

Answer


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Question 4:

Answer

Question 5:


Answer


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Question 6:

Answer

Question 7:

Answer



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Question 8:

Answer


Question 9:


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Answer

Question 10:

is equal to

A.

B.

C.


D.

Answer



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Question 11:

is equal to

A.

B.

C.

D.

Answer



Exercise 7.8Question 1:

Answer

It is known that,


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Question 2:

Answer

It is known that,


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Page 125 of 216Question 3:


Answer

It is known that,


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Question 4:

Answer

It is known that,


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Question 5:

Answer

It is known that,


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Question 6:

Answer

It is known that,



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Exercise 7.9

Question 1:

Answer

By second fundamental theorem of calculus, we obtain


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Question 2:

Answer


Question 3:


Answer



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Question 4:

Answer



Question 5:


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Question 5:

Answer


Question 6:

Answer



Question 7:


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Answer


Question 8:

Answer




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Question 9:

Answer


Question 10:


Answer


Question 11:


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Question 11:

Answer


Question 12:


Answer



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Question 13:

Answer



Question 14:

Answer


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Question 15:

Answer




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Question 16:

Answer

Let



A = 10 and B = 25


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Substituting the value of I1in (1), we obtain


Question 17:

Answer


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Question 18:

Answer




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Question 19:

Answer



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Question 21:

equals


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A.

B.

C.

D.

Answer




Question 22:

equals

A.

B.

C.

D.

Answer


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Answer





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Exercise 7.10

Question 1:

Answer

Whenx= 0, t= 1 and whenx= 1, t= 2


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Question 2:

Answer

Also, let



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Question 3:

Answer

Also, letx= tan dx= sec2d

Whenx= 0, = 0 and whenx = 1,


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Question 7:

Answer

Letx+ 1 = t dx= dt

Whenx= 1, t = 0 and whenx= 1, t = 2


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Question 8:

Answer

Let 2x=t 2dx= dt

Whenx= 1,t= 2 and whenx= 2, t= 4


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Let cot=t cosec2 d= dt



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Question 10:

If

A.cosx+xsinx

B.xsinx

C.xcosx

D. sinx +xcosx

Answer



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Exercise 7.11

Question 1:

Answer

Adding (1) and (2), we obtain


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Question 2:

Answer




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Question 3:

Answer




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Question 4:

Answer


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Question 6:


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Answer

It can be seen that (x 5) 0 on [2, 5] and (x 5) 0 on [5, 8].

Question 7:

Answer



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Question 8:

Answer



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Question 9:

Answer



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Question 10:

Answer




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Question 11:

Answer

As sin2 (x) = (sin (x))2= (sinx)2= sin2x, therefore, sin2x is an even function.


It is known that if f(x) is an even function, then

Question 12:


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Answer



Question 13:


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Answer

As sin7 (x) = (sin (x))7= (sinx)7= sin7x, therefore, sin2x is an odd function.

It is known that, if f(x) is an odd function, then

Question 14:

Answer

It is known that,


Question 15:

Answer


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Question 16:

Answer


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Question 17:

Answer

It is known that,



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Question 18:

Answer

It can be seen that, (x 1) 0 when 0 x 1 and (x 1) 0 when 1 x 4


Question 19:

Show that if fand gare defined as and


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Answer


Question 20:


The value of is

A. 0

B. 2

C.

D.1

Answer

It is known that if f(x) is an even function, then and

if f(x) is an odd function then


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if f(x) is an odd function, then


Question 21:

The value of is

A.2

B.

C.0

D.


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Miscellaneous Solutions

Question 1:

Answer


A + B C 0


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A+B C= 0

B+ C = 0

A= 1





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Question 2:

Answer



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Question 3:

[Hint: Put ]

Answer



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Question 4:

Answer



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Question 5:

Answer


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On dividing, we obtain

Question 6:


Answer


A+ B= 0

B + C= 5

9A+ C = 0




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Question 7:

Answer


Letx a = t dx= dt

Question 8:


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Answer

Question 9:

Answer


Let sinx= t cosx dx= dt

Question 10:

Answer


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Question 11:

Answer



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Question 12:

Answer

Letx4 =t 4x3dx= dt


Question 13:

Answer

Let ex= t exdx= dt


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Question 14:

Answer


Equating the coefficients ofx3,x2,x, and constant term, we obtain

A+ C= 0

B+ D= 0

4A+ C= 0

4B + D= 1




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Question 15:

Answer

= cos3

x sinxLet cosx=t sinx dx=dt


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Question 18:

Answer


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Question 19:

Answer



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Question 20:


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Answer



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Question 21:

Answer


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Question 24:

Answer


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Question 25:

Answer


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Whenx = 0, t = 0 and


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Question 27:

Answer



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When and when


Question 28:

Answer


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When and when

As , therefore, is an even function.

It is known that if f(x) is an even function, then


Question 29:

Answer


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Question 30:

Answer


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Question 31:

Answer



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Question 32:

Answer




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Question 33:


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Hence, the given result is proved.


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Question 35:

Answer



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Question 38:

Answer


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Question 39:

Answer



Let 1 x2= t 2xdx= dt


Question 40:

Evaluate as a limit of a sum.

A


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Answer

It is known that,



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Question 41:

is equal to

A.


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Question 43:

If then is equal to


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A.

B.

C.

D.

Answer



Question 44:

The value of is

A.1

B.0

C. 1


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D.

Answer





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Chapter 7 Integrals

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