Integral Calculus - Definite Integrals (Lecture-02)

If f(x) is a periodic function of period T; then

0nTf(x)dx=n0Tf(x)dx

Solved Examples

1. 010e{x}dx=

(a) 10(e1)

(b) 10

Show Answer

Solution:

f(x) is a periodic function of period 1 , then

01e{x}dx=1001ex[x]dx

=1001exdx=10(ex)01=10.(e1)

Answer: (a)

2. π/2199π/21+cos2xdx is equal to

(a) 502

(b) 1002

(c) 1502

(d) 2002

Show Answer

Solution:

π/2199π/21+cos2xdx=π/2199π/22cos2xdx=2π2199π2|cosx|dx=20100π|cosx|dx=10020π|cosx|dx=100220π2|cosx|dx=2002(sinx)0π2=2002(sinπ2sin0)=2002

Answer: (d)

3. Let f be a real valued function satisfying f(x)+f(x+6)=f(x+3)+f(x+9). Then xx+12f(t)dt is

(a) linear function of x

(b) exponential function of x

(c) a constant

(d) None of these

Show Answer

Solution:

f(x)+f(x+6)=f(x+3)+f(x+9) …………………(1)

Replace x by x+3

f(x+3)+f(x+9)=f(x+6)+f(x+12). …………………(2)

Adding (1) and (2) we get

f(x)=f(x+12)

f(x) is a periodic function of period 12

Let g(x)=xx+12f(t)dt

g1(x)=f(x+12)f(x)=0(f(x)=f(x+12))

g(x) is constant

Answer: (c)

4. The value of π/4nππ/4|sinx+cosx|dx is

(a) 2n

(B) 22n

(c) 42n

(d) None of these

Show Answer

Solution:

π/4nππ/42|sin(x+π4)|dx

Put x+π4=t

20nn|sint|dt=2n0π|sint|dt

=2n(cost)0π=2n(cosπ+cos0)=22n.

Answer: (b)

5. If ab|sinx|dx=8 and 0a+b|cosx|dx=9

then the value of a and b are

(a) π4,17π4

(b) 3π4,19π4

(c) π6,25π6

(d) None of these

Show Answer

Solution:

ba=4π & (a+b)0=9π2

Solving we get a=π4 b=17π4

6. If A=1sinθtdt1+t2 &  B=1cosecθdtt(1+t2), then the value of |AA2 BeAeBB211 A2+B21| is

(a) sinθ

(b) cosecθ

(c) 0

(d) 1

Show Answer

Solution:

Put t=1z in Bdt=1z2dz

the integral B reduces to

1sinθ1z2dz1z(1+1z2)=1sinθzdz1+z2=1sinθtdt1+t2=A

A+B=0

|AA2 BeAeBB211 A2+B21|=|AA2A1 A2112 A21|=0

(Applying C1C1+C3 )

7. e1tanxtdt1+t2+e1cosxdtt(1+t2)=

(a) 1

(b) e

(c) e1

(d) π4

Show Answer

Solution:

Put t=1z in 2nd integral

dt=1z2dz

2nd integral reduces to etanx1z2dz1z(1+1z2)=tanxezdz1+z2

1etanxtdt1+t2+tanxetdt1+t2=1cetdt1+t2

=12(loge(1+t2))1ee=12loge(1+e21+e2)=12logee2=1

Answer : (a)

Exercise

1. I=010sinxdx

(a) 10

(b) 10(1cos1)

(c) 5(1cos1)

(d) 5cos1

Show Answer Answer: b

2. π10π|sinx|dx=

(a) 20

(b) 8

(c) 10

(d) 18

Show Answer Answer: d

3. If I1=03πf(cos2x)dx & I2=0πf(cos2x)dx then

(a) I1=I2

(b) I1=2I2

(c) I1=5I2

(d) I1=3I2

Show Answer Answer: d

4. limn(n!)1nnn is equal to

(a) e

(b) 1e

(c) e-1

(d) e5

Show Answer Answer: b

5. The value of the definite integral 01xdxx3+16 lies in the interval [a,b]. Then smallest such interval is

(a) [0,117]

(b) [0,1]

(c) [0,127]

(d) [117,1]

Show Answer Answer: a

6. Let f:RR such that f(x+2y)=f(x)+f(2y)+4xy,x,yR & f(0)=0 If I1=01f(x)dx,

I2=01f(x)dx & I3=122f(x)dx, then

(a) I1=I2>I3

(b) I1>I2>I3

(c) I1= I2<I3

(d) I1< I2< I3

Show Answer Answer: c

7. The integral tan1λcot1λtanxtanx+cotxdx,λR can not take the value

(a) π4

(b) π2

(c) π4

(d) 3π4

Show Answer Answer: a, b, d

8. Read the following and answer the questions that follow :-

I=010πcos6xcos7xcos8xcos9x1+e2sin34xdx

(i) If I=k limits0π2cos6xcos7xcos8xcos9xdx, then k=

(a) 5

(b) 10

(c) 20

(d) None

(ii) If I=c0π2cos6xcos8xcos2xdx, then c=

(a) 5

(b) 10

(c) 20

(d) None

(iii) The value of I=

(a) 5π4

(b) 5π8

(c) 5π16

(d) 5π12

Show Answer Answer: (i) b (ii) a (iii) b

9. Match the following if

I1=01etan1x1+x2dx,I2=01xtan1x1+x2dx,I3=01x2etan1x1+x2dx, then

Column I Column II
(a) I1+I2 (p) 3/2
(b) I1+I24I3 (q) 1
(c) I1+I2+1I1+I2+2I3 (r) 2eπ/41
(d) 2(I1+I2)2I3+1 (s) 2eπ/4
(t) 2eπ/43
Show Answer Answer: ar,bt,cq,dr

10. Let Sn=k=0nnn2+kn+k2 & Tn=k=0n1nn2+kn+k2 for n=1,2,3 then

(a) Sn<π33

(b) Sn>π33

(c) Tn<π33

(d) Tn>π33

Show Answer Answer: a, d

11. 0[x]2x2[x]dx is

(a) [x]loge2

(b) [x]loge2

(c) 12[x]loge2

(d) None of these.

Show Answer Answer: b

12. The number of positive continuous functions f(x) defined in [0,1] for which 01f(x)dx=1,

01xf(x)dx=a,01x2f(x)dx=a2 is

(a) one

(b) infinite

(c) two

(d) zero

Show Answer Answer: d


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