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RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:20 PM IST

The class 12 RD Sharma chapter 19 exercise MCQ solution deals with the chapter of definite integrals which is a sub part of the indefinite integrals chapter, The Definite integral is defined to be the exact limit and summation that we looked at in the past section to find the net area between a function and the X-axis . The RD Sharma class 12th exercise MCQ gives us the brief practice of the entire chapter and makes us go through all the concepts one by one to revise each and every topic of the chapter thoroughly.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter19 MCQ Definite Integrals - Other Exercise

Definite Integrals Excercise: MCQ

Definite Integrals exercise Multiple choice question 1

Answer:
π8
Hint:
Using a2x2dx
Explanation:
01x(1x)dx=01xx2dx
=01(x)2x+(12)2(12)2dx=01(x12)2+(12)2dx=01(12)2(x12)2dx
We know
a2x2=x2a2x2+a22sin1xa+c
I=[x1/2214(x12)2+18sin1(x1/22)]01
I=[2x1414(x12)2+18sin1(x1/21/2)]01
=[2x1414(x12)2+18sin1(2x1)]01
=(1414(112)2+18sin1(1))(1414(012)2+18sin1(1))
=(141414+18×π2)(141414+18×(π2))=0+π160(π16)=π8

Definite Integrals exercise Multiple choice question 2

Answer:
2
Hint:
Using xdx and ddx(tanx)
Given:
0π11+sinx
Explanation:
Let
I=0π11+sinx=0π11+2tanx/21+tan2x/2dx
=0π1+tan2x21+tan2x/2+2tanx/2dx=sec2x2tan2x/2+2tanx2+1dx

 Put 2tanx2=tsec2x/21/2dx=dtsec2x/2dx=2dt
When x=0 then t=0
When x=π then t=
=02dtt2+2t+1=01(t+1)2dt
=0(t+1)2dt=[(t+1)22+1]9
=2[1(t+1)]9=2(01)=2

Definite Integrals exercise Multiple choice question 3

Answer:
π24
Hint:
0af(x)dx=0af(ax)dx
Given:
0πxtanxsecx+cosxdx
Explanation:
Let
I=0πxtanxsecx+cosxdx,[0af(x)dx=0af(ax)dx]
=0π(πx)tanxsecxcosxdx
I=0π(πx)tanxsecx+cosxdxI=0ππtanxsecx+cosxdx0πxtanxsecx+cosxdx
I=π0πtanxsecx+cosxdxI2I=π0πsinxcosx1cosx+cosxdx2I=π0πsinx1+cos2xdx
Put
cosx=tsinxdx=dt
When x=0 then t=1
When x=π then t=1
=π11dt1+t2
=π211dt1+t2,[baf(x)dx=abf(ax)dx]
=π2[tan1t]11=π2[tan1(1)tan1(1)]=π2[π4(π4)]
=π2(π2)=π24


Definite Integrals exercise Multiple choice question 4

Answer:
4
Given:
02π1+sinx/2dx
Hint:
Using sinxdx,cosxdx
Explanation:
Let
I=02π1+sinx/2dx
[sin2θ=2sinθcosθ,sin2θ+cos2θ=1]=02π(sinx/4+cosx/4)2dx=02π(sinx/4+cosx/4)dx
=02πsinx/4dx+02πcosx/4dx=[4cosx4]02π+[4sinx4]02π
=4cos[x/4]02π+[4sinx4]02π=[4cosπ2+4cos0]+[4sinπ24sin0]
=0+4+0=4

Definite Integrals exercise Multiple choice question 5

Answer:
π4
Given:

0π2cosxcosx+sinxdx
Hint:
using 0af(x)dx=0af(ax)dx
Explanation:
Let
I=0π2cosxcosx+sinxdx(i)
I=0π2cos(π2x)cos(π2x)+sin(π2x)dx,[0af(x)dx=0af(ax)dx]=0π2sinxsinx+cosxdx(ii)
Adding (i) and (ii)
2I=0π2cosx+sinxsinx+cosxdx=0π21dx2I=[x]0π22I=[π20]
2I=π2I=π4

Definite Integrals exercise Multiple choice question 6

Answer:
log2
Given:

011+exdx
Hint:
Explanation:
Let
I=011+exdx=01ex(1ex+1)dx=0ex1+exdx
Put
1+ex=texdx=dtexdx=dt=0dtt=[log|t|]0
=[log|1+ex|]0=[log|1+e|][log|1+e0|]
=[log|1+0|][log|1+1|],[e=0,e0=1]=[0log2]=log2

Definite Integrals exercise Multiple choice question 7

Answer:
2
Given:
0π24sinxxdx
Hint:
Using sinxdx
Explanation:
Let
I=0π24sinxxdx Put x=t12xdx=dt1xdx=2dt
=0π2sint2tdt=2[cost1]0π2
=2[cosπ2+cos0]=2[0+1]=2

Definite Integrals exercise Multiple choice question 8

Answer:
log(43)
Hint:
You must know about log properties and using partial fraction.
Given:
0π2cosx(2+sinx)+(1+sinx)dx
Explanation:
Let
I=0π2cosx(2+sinx)(1+sinx)dxI=0π2cosx(1+1+sinx)(1+sinx)dx
Put
1+sinx=tcosxdx=dt
When x=0 then t=1
When x=π2 then t=2
=02dt(t+1)t1(t+1)t=At+Bt+1
Multiplying by t(t+1)
1=A(t+1)+B(t)
Putting t=1
1=A(0)+B(1)
B=1
Putting t=1
1=A(1)+B(0)A=1
1(t+1)t=1t1t+102dt(t+1)t=021t1t+1dt
=021tdt021t+1dt=[log|t|]12[log|t+1|]12=[log2log1][log3log2]
=log20log3+log2=2log2log3=log22log3=log4log3=log(43)

Definite Integrals exercise Multiple choice question 9

Answer:
23tan1(13)
Given:
0π212+cosxdx
Hint:
Using11+x2dx

Explanation:
Let
I=0π212+cosxdx
Put
cosx=1tan2x/21+tan2x/2=0π212+1tan2x/21+tan2x/2dx
=0π21+tan2x/22(1+tan2x/2)+1tan2x/2dx
=0π2sec2x/23+tan2x/2dx
 Put tanx/2=tsec2x/21/2dx=dtsec2x/2dx=2dt
When x=0 then t=0
When x=π2 then t=1
=012dtt2+(3)2=2[13tan1(t3)]01=23[tan1(13)tan1(0)]
=23[tan1(13)0]=23tan1(13)

Definite Integrals exercise Multiple choice question 10

Answer:
π21
Given:
011x1+xdx
Hint:
Explanation:
Let
I=011x1+xdx=011x1+x×1x1xdx
=01(1x)21x2dx=011x1x2dx
=0111x2dx01x(1x2)1/2dx=0111x2dx+01(1x2)1/2(2x)dx
=[sin1x]01+12[(1x2)1/21/2]01=[sin1(1)sin1(0)]+12[011/2]
=(π20)+12(2)=π21

Definite Integrals exercise Multiple choice question 11

Answer:
πa2b2
Given:
0π1a+bcosxdx
Hint:
Using11+x2dx
Explanation:

Let
I=0π1a+bcosxdxcosx=1tan2x/21+tan2x/2
=0π1a+b(1tan2x/21+tan2x/2)dx=0π1+tan2x/22a+atan2x/2+bbtan2x/2dx
=0πsec2x2(ab)tan2x/2+(a+b)dx Put tanx/2=tsec2x/21/2dx=dtsec2x/2dx=2dt
=02dt(ab)t2+(a+b)
=2(ab)0dtt2+(a+bab)2
=2ab{tan1[taba+b]}0×1a+bab
=2ababa+b(tan1tan10)=2(a+b)(ab)(π2)=πa2b2

Definite Integrals exercise Multiple choice question 12

Answer:
π12
Given:
π6π311+cotxdx
Hint:
Usingabf(x)dx=abf(a+bx)dx

Explanation:
Let
I=π6π311+cotxdxI=π6π311+cosxsinxdx,[cotx=cosxsinx]
=π6π3sinxsinx+cosxdx(i)
We have a property that
abf(x)dx=abf(a+bx)dx
Using this, we get
=π6π3cos(π3+π6x)cos(π3+π6x)+sin(π3+π6x)dx
=π6π3cosxcosx+sinxdx,[sin(π2x)=cosx,cos(π2x)=sinx]
I=π6π3cosxcosx+sinxdx(ii)
Adding (i) and (ii)
2I=π6π3sinx+cosxsinx+cosxdx=π6π31dx
I=12[x]π6π3I=12[π3π6]
I=12[2π2π6]I=12×π6=π12

Definite Integrals exercise Multiple choice question 13

Answer:
π60
Given:
0x2(x2+a2)(x2+b2)(x2+c2)dx=π2(a+b)(b+c)(c+a)
Hint:
Using given condition find a,b,c.

Explanation:
Given that
0x2(x2+a2)(x2+b2)(x2+c2)dx=π2(a+b)(b+c)(c+a)(i)
We have to evaluate
0dx(x2+4)(x2+9)
Let
I=0dx(x2+4)(x2+9)
Multiply and divide by x2
I=0x2dx(x2+4)(x2+9)(x2+0)
On comparing with equ (i)
We get
a2=4;b2=9;c2=0a=2;b=3;c=0
To given
0x2(x2+a2)(x2+b2)(x2+c2)dx=π2(a+b)(b+c)(c+a)
0x2(x2+a2)(x2+b2)(x2+c2)dx=π2(2+3)(3+0)(0+2)
=π2×5×3×2
=π60

Definite Integrals exercise Multiple choice question 14

Answer:
1
Given:
1θlogxxdx
Hint:
Using by parts.
Explanation:
Let
I=1elogxdx=[logx1dxddx(logx)dx]1e
=[xlogx1xxdx]1e=[xlogxx]1e

=[x(logx1)]1e=1[01]=1

Definite Integrals exercise Multiple choice question 15

Answer:
π12
Given:
1311+x2dx
Hint:
Use formula 11+x2dx=tan1x

Explanation:
I=1311+x2dx=[tan1x]13=tan13tan11
=π3π4=4π3π12=π12

Definite Integrals exercise Multiple choice question 16

Answer:
π12+log2(2)
Given:
033x+1x2+9dx
Hint:
Usingdxx and 11+x2dx

Explanation:
Let
I=033x+1x2+9dx=033xx2+9dx+031x2+9dx
=033xx2+9dx+031x2+9dx=[32log|x2+9|+13tan1x3]03
=32log18+13tan1132log9+13tan10=32(log18log9)+13tan1(tanπ4)+13tan1(tan0)
=32(log189)+13×π4+13×0,[logmlogn=logmn]=32log2+π12+0
=log(2)3/2+π12=π12+log2(2)


Definite Integrals exercise Multiple choice question 17

Answer:
π4
Given:
0x(1+x)(1+x2)dx
Hint:
You must know derivation of sinx, cosx and tanx.

Explanation:
Let
I=0x(1+x)(1+x2)dx=0π2tanθ(1+tanθ)(1+tan2θ)sec2θdθ
=0π2tanθ(1+tanθ)(sec2θ)sec2θdθ=0π2tanθ+111+tanθdθ
=0π21dθ0π211+tanθdθ=0π21dθ0π211+sinθcosθdθ,[tanθ=sinθcosθ]
=0π21dθ0π2cosθcosθ+sinθdθ =[θ]0π2I1, where I1=0π2cosθcosθ+sinθdθ
=(π20)I1=π2I1(i)
Now,
I1=0π2cosθcosθ+sinθdθ(ii)=0π2cos(π2θ)cos(π2θ)+sin(π2θ)dθ
=0π2sinθsinθ+cosθdθ,[0af(x)dx=0af(ax)dx]I1=0π2sinθsinθ+cosθdθ(iii)
Adding (ii) and (iii)
I1=0π2cosθ+sinθsinθ+cosθdθ=0π21dθ
=12[θ]0π2=12[π20]=π4
Put in (i)
I=π2π4I=2ππ4I=π4

Definite Integrals exercise Multiple choice question 18

Answer:
2
Given:
π2π2sin|x|dx
Hint:
You must know about |x| function and sinxdx

Explanation:
Let
I=π2π2sin|x|dx
We know that
|x|={x,x0x,x<0I=π20sin(x)dx+0π2sinxdx
I=π20sinxdx+0π2sinxdx=[cosx]π20+[cosx]0π2
=[cosx]π20[cosx]0π2=cos(0)cos(π2)cosπ2+cos(0)=100+1=2


Definite Integrals exercise Multiple choice question 19

Answer:
π4
Given:
0π211+tanxdx
Hint:
using0af(x)dx=0af(ax)dx

Explanation:
Let
I=0π211+tanxdxI=0π211+sinxcosxdx
I=0π2cosxcosx+sinxdx(i)I=0π2cos(π22)cos(π22)+sin(π22)dx
I=0π2sinxsinx+cosxdx(ii)
Adding (i) and (ii)
2I=0π2sinx+cosxsinx+cosxdx2I=0π21dx
I=12[x]0π2I=12[π20]=π4

Definite Integrals exercise Multiple choice question 20

Answer:
e1
Given:
0π2cosxesinxdx
Hint:
You must know about ddx(sinx) and etdt

Explanation:
Let
I=0π2cosxesinxdx
Put
sinx=tcosxdx=dt
Whenx=0 then t=0
When x=π2 then t=1
=01etdx=[et]01=e1e0=e1

Definite Integrals exercise Multiple choice question 21

Answer:
12
Given:
0a11+4x2dx=π8
Hint:
To solve this equation, we will do 4x in whole system.

Explanation:
Let
I=0a11+4x2dxI=0a11+(2x)2dx
=[12tan1(2x1)]0a=π812[tan1(2a)tan1(0)]=π8
12[tan1(2a)]=π8tan1(2a)=π4
2a=tanπ42a=1a=12

Definite Integrals exercise Multiple choice question 22

Answer:
0
Given:
01f(x)dx=1;01xf(x)dx=a;012x2f(x)dx=a2
Hint:
To solve this equation, we have to integrate differentially.

Explanation:
01f(x)dx=1(i)01xf(x)dx=a(ii)012x2f(x)dx=a2(iii)
I=01(ax)2f(x)dxI=01(a2+x22ax)f(x)dx
I=01a2f(x)dx+01x2f(x)dx012axf(x)dxI=a201f(x)dx+01x2f(x)dx2a01xf(x)dx
I=a2+a22(a)(a)I=2a22a2I=0

Definite Integrals exercise Multiple choice question 23

Answer:
0
Given:
ππsin3xcos2xdx
Hint:
This equation is solve by aaf(x)dx

Explanation:
aaf(x)dxf(x)=sin3(x)cos2(x)
=(sinx)3cos2x=sin3xcos2x=f(x)
f(x) is odd function.
ππsin3xcos2xdx=0

Definite Integrals exercise Multiple choice question 24

Answer:
I=log3
Given:
π6π31sin2xdx
Hint:
To solve this equation we convert sin into cosec
Explanation:
Let
I=π6π31sin2xdx=π6π3cosec2xdx
I=12[log|cosec2x+cot2x|]π6π3I=12[log(cosec2π3+cot2π3)][log(cosecπ3+cotπ3)]
I=12[log|2313|][log|23+13|]I=12[log13log3]
I=12[log3log3]I=12×2log3I=log3

Definite Integrals exercise Multiple choice question 25

Answer:
2
Given:
11|1x|dx
Hint:
To solve this equation we use xnn+1 formula.

Explanation:
Let
11|1x|dx=[xx22]11
=[112][112]=112+1+12=2

Definite Integrals exercise Multiple choice question 26

Answer:
(logx)1(x1)x
Given:
x2x31logetdx
Hint:
This equation will solve byf(x)formula.

Solution:
f(x)=1logx33x21logx2(2x)f(x)=3x23logx32x2logx2
f(x)=x2logxxlogx=x2xlogx
=x(x1)logx=(logx)1(x1)x



Definite Integrals exercise Multiple choice question 27

Answer:
10(π2)9
Hint:
To solve this equation we convert sin into cos.
Given:
I10=0π2x10sinxdx
Solution:
I10=0π2x10sinxdx=x10cosx0π210x9cosxdx
=0100π2x9cosxdx=10[x9sinx90π2x8sinxdx]0π2
=10(π2)9900π2x8sinxdx=10(π2)990I8=10(π2)9

Definite Integrals exercise Multiple choice question 28

Answer:
12756
Hint:
Integrate then use limits
Given:
01x(1x)54dx
Solution:
I=01x(1x)54dxI=01x(x1)54[Q(xy)2=(yx)2]
=01x+11(x1)54dx=01(x1)(x1)54dx+011(x1)54dx
=01(x1)53dx+01(x1)54dx=[(x1)5252]01+[(x1)5353]01
=[(x1)5252]01+[(x1)5353]01=(1)5252+(1)5353
=152153=152×53=12756


Definite Integrals exercise Multiple choice question 29

Answer:
2(21)
Hint:
First integrate and then use limits
Given:
0π21sin2xdx
Solution:
I=0π21sin2xdxI=0π2|sinxcosx|2dx
I=0π2|sinxcosx|dx=π4π2(sinxcosx)dx0π4(sinxcosx)dx
I=[cosxsinx]π4π2|cosxsinx|0π4I=01+2(2+1)=222I=2(21)

Definite Integrals exercise Multiple choice question 30

Answer:
4
Hint:
To solve this we will split 1x2 in differential form.
Given:
22|1x2|dx
Solution:
22|1x2|dx1x2=0x2=1x=±1
=21(1x2)dx+11(1x2)dx+12(1x2)dx=[x33x]21+[xx33]11+[x33x]12
=[13+1(83)+2]+[113(1)+13]+[832(131)]=23(23)+[23(23)]+[23+23]
=43+43+43+123=4


Definite Integrals exercise Multiple choice question 31

Answer:
π4
Hint:
To solve this equation we should simplify cot x.
Given:
0π211+cotxdx
Solution:
Let
0π211+cotxdx
I=0π211+cosxsinxdx,[cotx=cosxsinx]
=0π2sinxcosx+sinxdx(i),[0af(x)dx=0af(ax)dx]
=0π2sin(π2x)cos(π2x)+sin(π2x)dx
=0π2cosxcosx+sinxdx(ii)
Adding (i) and (ii)
2I=0π2cosx+sinxcosx+sinxdx2I=0π21dx
I=12[x]0π2I=12[π20]=π4


Definite Integrals exercise Multiple choice question 32

Answer:
π4
Hint:
To solve this equation we use baf(x)dx formula.
Given:
0π2sinxsinx+cosxdx
Solution:
Let
I=0π2sinxsinx+cosxdx(i)0π2f(x)dx=0π2f(π2x)dx
=0π2sin(π2x)sin(π2x)+cos(π2x)dx=0π2cosxcosx+sinxdx(ii)
Adding (i) and (ii)
2I=0π2sinxsinx+cosxdx+0π2cosxcosx+sinxdx=0π21sinx+cosx[cosx+sinx]dx
2I=0π21dx2I=[x]0π2
2I=[π20]=π2×12I=π4

Definite Integrals exercise Multiple choice question 33

Answer:
π2
Hint:
To solve this equation we use abddxf(x)dx formula.
Given:
01ddx{sin1(2x1+x2)}dx
Solution:
Let
I=01ddx{sin1(2x1+x2)}dx=01ddx(2tan1x)dx
=201ddx(tan1x)dx=20111+x2dx
=2[tan1x]01=2[tan11tan10]=2[π4]0=π2

Definite Integrals exercise Multiple choice question 34

Answer:
1
Hint:
To solve this equation we suppose sin x as cos x.
Given:
0π2xsinxdx
Solution:
Let
I=0π2xsinxdx
=[xsinxdx+cosxdx]0π/2=[xcosx+sinx]0π/2
=π2cosπ2+sinπ20=π2cosπ2+1=0+1=1

Definite Integrals exercise Multiple choice question 35

Answer:
0
Hint:
To solve this equation we use abf(x)dx formula.
Given:
0π2sin2xlogtanxdx
Solution:
Let
I=0π2sin2xlogtanxdx(i)abf(x)dx=abf(a+bx)dx
I=0π2sin2(π2x)log[tan(π2x)]dx=0π2sin(π2x)logcotxdx
=0π2sin2xlogcotxdx(ii)
Adding (i) and (ii)
2I=0π2sin2xlogtanx+sin2xlogcotxdx
2I=0π2sin2x(logtanx+logcotx)dx2I=0π2sin2x(logtanxcotx)dx
logab=loga+logb2I=0π2sin2x(logtanx1tanx)dx
2I=0π2sin2x(log1)dx2I=0I=0


Definite Integrals exercise Multiple choice question 36

Answer:
π4
Hint:
To solve this equation we convert cos in term of tan.
Given:
0π15+3cosxdx
Solution:
Let
I=0π15+3cosxdxcos2θ=1tan2θ1+tan2θ
cosθ=1tan2θ/21+tan2θ/2
I=0π15+3(1tan2x21+tan2x2)dx
I=0πsec2x28+2tan2x2dx Let tanx2=t,sec2x212dx=dt
=220dt4+t2=01(2)2+t2dt
=[12tan1(t2)]0=12(π2)=π4

Definite Integrals exercise Multiple choice question 37

Answer:
πlog2
Hint:
To solve this equation we suppose x as tan x.
Given:
0log(x+1x)(11+x2)dx
Solution:
Let
I=0log(x+1x)(11+x2)dxx=tanx
dx=sec2θdθ When x=0,tanθ=0,θ=0 When x=,tanθ=,θ=π2
I=0π2log(tanθ+cotθ)sec2θdθ1+tan2θ=0π2log(sinθcosθ+cosθsinθ)dθ
=0π2sin2θ+cos2θsinθcosθdθ=0π2log1log(sinθcosθ)dθ
=0π2logsinθdθ0π2logcosθdθ=(π2log2)(π2log2)
I=0

Definite Integrals exercise Multiple choice question 38

Answer:
a2af(x)dx=0af(x)dx+02af(2ax)dx
Hint:
Given:
a2af(x)dx
Solution:
a2af(x)dx
f(2ax)=f(x)0a2f(x)dx
=20af(x)dxa2af(x)dx=0af(x)dx+02af(2ax)dx

Definite Integrals exercise Multiple choice question 39

Answer:
I=a+b2abf(x)dx
Hint:
To solve this we will split (a+b-x)
Given:
(a+bx)=f(x)
Solution:
(a+bx)=f(x)
Let
I=abxf(x)dx(i)abf(x)dx=ab(a+bx)dx
I=ab(a+bx)f(a+bx)dxI=ab(a+bx)f(x)dx
I=ab(a+b)f(x)dxabxf(x)dx(ii)
Adding equation (i) and (ii)
I+I=abxf(x)dx+ab(a+b)f(x)dxabxf(x)dx
2I=ab(a+b)f(x)dxI=12ab(a+b)f(x)dxI=a+b2ab(a+b)f(x)dx

Definite Integrals exercise Multiple choice question 40

Answer:
0
Hint:
Given:
01tan1(2x11+xx2)dx
Explanation:
I=01tan1(2x11xx2)dx=01tan1[x(1x)1+x(1x)]dx
=01tan1xtan1(1x)dx .........Eq(i)
Using properties,
0af(x)dx=0af(ax)dxI=01tan1(1x)tan1(1(1x))dx
=01tan1(1x)tan1xdx ..........Eq(ii)
Adding Eq(i) and Eq(ii)
2I=0I=0

Definite Integrals exercise Multiple choice question 41

Answer:
I=0
Hint:
To solve this equation we use 0af(x)dx formula.
Given:
0π2log[4+3sinx4+3cosx]dx
Solution:
Let
I=0π2log[4+3sinx4+3cosx]dx(i)0af(x)dx=0a(ax)dx
I=0π2log[4+3sin(π2x)4+3cos(π2x)]dxI=0π2log[4+3cosx4+3sinx]dx(ii)
Adding (i) and (ii)
2I=0π2(log[4+3cosx4+3sinx]+log[4+3sinx4+3cosx])dx2I=0π2(log[4+3cosx4+3sinx]log[4+3sinx4+3cosx])dx
2I=0π21dx2I=0I=0

Definite Integrals exercise Multiple choice question 42

Answer:
π
Hint:
We use aaf(x)dx=0 function.
Given:
π2π2(x3+xcosx+tan5x+1)dx
Explanation:
aaf(x)dx=0f(x) odd function f(x)=f(x)(x)3=x3(i)
xcosx(x)cos(x)=xcosx(ii)tan5xtan5(x)=tan5x(iii)
=π2π2x3dx+π2π2xcosxdx+π2π2tan5xdx+π2π21dx=π2π21dx
=[x]π2π2=π2(π2)=2π2=π

Definite Integrals exercise Multiple choice question 43

Answer:
1
Hint:
To solve this equation.
Given:
π4π411+cos2xdx
Explanation:
π4π411+cos2xdx=π4π412cos2xdx1+cos2x=2cos2x
=12π4π41cos2xdx=12π4π4sec2xdx1cosx=secx
=12[tanx]π4π4=12[tanπ4tan(π4)]=12[1(1)]
=12[1+1]=22=1

Definite Integrals exercise Multiple choice question 44

Answer:
abf(x+c)dx
Hint:
To solve this abf(x)dx formula is used.
Given:
a+cb+cf(x)dx
Explanation:
a+cb+cf(x)dx
=f[x]a+cb+c=f(b+c)f(a+c)
Now, solving option (b),
abf(x+c)dx=[f(x+c)]ab=f(b+c)f(a+c)
Hence, option (b) is correct.

Definite Integrals exercise Multiple choice question 45

Answer:
a20af(x)dx
Hint:
To solve this equation we use 0af(x)dx formula.
Given:
f(x)=f(ax),g(x)+g(ax)=a
Explanation:
I=0ag(x)f(x)dx(i) Put x=(ax)=0ag(ax)f(ax)dx(ii)
=0af(x)(ag(x))dxI=a0af(x)dx0af(x)g(x)dx
I=a0af(x)dxI2I=a0af(x)dxI=a20af(x)dx

Definite Integrals exercise Multiple choice question 46

Answer:
1
Hint:
To solve this question we have to use substitution method.
Given:
01tan(sin1x)dx
Explanation:
sin1x=tx=sintdx=costdt When x=0,sin10=0 When x=1,sin11=π2
I=0π2tantcostdtsintcost=tantsint=costcost
=0π2sintdtI=[cost]0π2I=0(1)=1


The RD Sharma class 12 solution of Definite integrals exercise MCQ covers the chapter 'Definite Integrals.' There are about 46 MCQs in this exercise which gives a brief insight of the chapter. The RD Sharma class 12th exercise MCQ covers all the essential concepts like,

  • Limit sum of definite integrals

  • Fundamentals of a theorem in definite integral

  • Basic properties

  • Evaluation method of integrals

The beneficial factors of using the RD Sharma class 12 solution chapter 19 exercise MCQ are :-

  • RD Sharma class 12th exercise MCQ is available for download on the careers360 website, you can visit the site for free PDFs and study materials.

  • Any student who has previously used the RD Sharma class 12 chapter 19 exercise MCQ solution for practicing will always raise the study material of RD Sharma as it has always helped them to score better and to understand maths easily.

  • The RD Sharma class 12th exercise MCQ contains questions that are prepared by maths experts with some helpful tips to solve the questions easily.

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  • Every student should make use of the RD Sharma class 12th exercise MCQ to stay ahead of the class as most of the Teacher's take reference of the solution to provide lectures and prepare question papers.

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Frequently Asked Questions (FAQs)

1. Mention some definite rephrasing rules?

 The integral of a sum is said to be  the sum of the integrals whereas the integral of a difference is said to be the difference of the integrals.Similarly, we can say that  the integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function.

2. What is a definite integral used for?

definite integrals can be used to determine the mass of an object if its density function is known

3. Can a definite integral be negative?

expressed more compactly,  The Definite integral is the sum of the areas above minus the sum of the areas below, therefore area is always non negative.

4. Is Definite integral important for exams?

Yes,  The Definite integral chapter is important for exams as most of the questions asked for higher marks are from these chapters

5. Can the RD Sharma solutions be used for self practice?

Yes, you can definitely use the RD Sharma solutions for self practice and then you can evaluate the performance according to the  answers given in the book.

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