Integral Calculus Definite Integrals(Lecture-03)

Property:

Limit as a sum

  • Express the given series in the form of 1nf(rn)

  • Then the limit as its sum when n i.e. limn1n.f(rn)

  • Replace rn by x and 1n by dx and 1n by the sign of

Solved Examples

1. limn1n+1+1n+2+1n+3++12n=

(a) 1

(b) 2

(c) loge2

(d) loge4

Show Answer

Solution: limnr=1n1n+r=limn1nr=1n11+rn

=01dx1+x=loge2

Answer: c

2. The value of r=0n114n2r2 as n is

(a) π3

(b) π6

(c) π4

(d) none of these

Show Answer

Solution: r=0n114n2r2=1nr=0n114(rn)2=01dx4x2=(sin1x2)01

=sin112sin10=π6

Answer: b

3. The value of limn(12+22+32++n2)(13+23++n3)16+26++n6

(a) 1

(b) 712

(c) 56

(d) none

Show Answer

Solution: limn(r=1nr2r=1nr3)r=1nr6

=limn(1nr=1n(rn)2)(1nr=1n(rn)3)1nr=1n(rn)6

=01x2dx01x3dx01x6dx=712

4. 01[2ex]dx is equal to

(a) loge2

(b) e2

(c) 0

(d) 2e

Show Answer

Solution: 02[2ex]dx=0logc21dx+logc20dx

=loge2

Answer: a

5. The value of 0π/3[3tanx]dx is

(a) 5π6

(b) 5π6tan123

(c) π2tan123

(d) none

Show Answer

Solution: 0π/3[3tanx]dx=0π/60dx+π/6tan1231dx+tan123π/32dx

=0+(tan123π6)+2(π3tan123)=π2tan123

Answer: c

6. 11[x[1+sinπx]+1]dx is equal to

(a) 2

(b) 0

(c) 1

(d) none

Show Answer

Solution: 101dx+01[x+1]dx

=10[x0+1]dx+01[x1+1]dx

=101dx+01([x]+1)dx=(x)10+(x)01=1+1=2

Answer: a

7. 02[x21]dx is equal to

(a) 332

(b) 2

(c) 1

(d) 32332

Show Answer

Solution: 01(1)dx+120dx+231dx+322dx

=(10)+0+1(32)+2(23)=1+32+423=332

Answer: a

Exercise

1. limn1P+2P+3P++nPnP+1=

(a) 1P+1

(b) 11P

(c) 1P1P1

(d) PP+1

Show Answer Answer: a

2. Let a,b,c be non-zero real numbers such that 01(1+cos8x)(ax2+bx+c)dx

=02(1+cos8x)(ax2+bx+c)dx. Then the quadratic equation ax2+bx+c=0 has

(a) no root in (0,2)

(b) at least one root in (1,2)

(c) a double root in (0,2)

(d) two imaginary roots

Show Answer Answer: b

3. For every function f(x) which is twice differentiable these will be good approximation of

abf(x)dx=ba2(f(a)+f( b)), for more accurate results for c(a,b)F(c)=ca2(f(a)f(c))+bc2(f( b)f(c)) where c=a+b2abf(x)dx=ba4(f(a)+f( b)+2f(c))dx

i. Good approximation of 0π/2sinxdx is

(a) π4

(b) π(2+1)4

(c) π(2+1)8

(d) π8

ii. If f(x)<0,x(a,b) & (c,f(c)) is point of maximum where c(a.b) then f(c) is

(a) f( b)f(a)ba

(b) 3(f( b)f(a)ba)

(c) 2(f( b)f(a)ba)

(d) 0

iii. If limtaatf(x)dxta2(f(t)+f(a))(ta)3=0, then degree of polynomial function f(x) at most is

(a) 0

(b) 1

(c) 3

(d) 2

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

4. Match the following:-

Column I Column II
a. 0e4xsin5xdx (p) 3
b. 28[x2]dx[x220x+100]+[x2] (q) 541
c. 03π/2|sinx|dx,nN (r) 120
d. 0x5exdx (s) 60
Show Answer Answer: aq;bp;cp;dr;

5. 13dxx2+[x]2+12x[x]=

(a) π

(b) 3π

(c) 8π

(d) π2

Show Answer Answer: d

6.* If In=ππsinnx(1+πx)sinxdx,n=0,1,2, then

(a) In=In+2

(b) m=110I2 m+1=10π

(c) m=110I2 m=0

(d) In=In+1

Show Answer Answer: a, b, c

7. All the values of a for which 12(a2+(44a)x+4x3)dx12 are given by

(a) a4

(b) a=3

(c) 0a<3

(d) none of these

Show Answer Answer: b

8. Value of 02[x2x+1]dx is

(a) 7+52

(b) 552

(c) 752

(d) none of these

Show Answer Answer: b

9. If I=01dx(5+2x2x2)(1+e24x) and I1=01dx5+2x2x2dx then

(a) I=12I1

(b) I=2I1

(c) 3I=5I1

(d) none of these

Show Answer Answer: a

10. Match the following:-

Column I Column II
a. limnr=1nr2r3+n3 (p) loge2
b. limn1n(tanπ4n+tan2π4n+..+tanπ4) (q) 2loge2
c. limn1n(1+1n2)2/n2(1+22n2)4/n2..(1+n2n2)2n/n2 (r) 2πloge2
d. 0[2ex]dx (s) 13loge2
Show Answer Answer: as;br;cq;dp;

11. 02[tan1x]dx+02[cot1x]dx=

(a) 1cot2

(b) 1+cot2

(c) 2(1+cot2)

(d) 2(1cot2)

Show Answer Answer: c

12.* Match the following

([.] denotes the greatest integer function)

Column I Column II
a. If μ<01x7dx1+x83<λ, then (p) [λ+μ]=2
b. If μ<01dx1+x6<λ, then (q) [λ+μ]=4
c. If μ<01dx4x2x3<λ, then (r) [λμ]=0
(s) [λμ]=3
(t) [λ+μ]=0
Show Answer Answer: ar,t;bp,r;cq,s;