Definite Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Definite Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 24, 2025

നേടുക Definite Integrals ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Definite Integrals MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Definite Integrals MCQ Objective Questions

Top Definite Integrals MCQ Objective Questions

Definite Integrals Question 1:

The following plot shows a function y which varies linearly with x. The value of I=12ydx is ________.

F1 Shubham Shraddha 09.12.2020 D15

Answer (Detailed Solution Below) 2.49 - 2.51

Definite Integrals Question 1 Detailed Solution

Given:

I=12ydx

Since the integration is done w.r.t 'x' variable, 'y' in terms of x can be written as:

y = mx + c

The slope of the given line will be:

m=100(1)=1

y = x + c

when x = 0, y = 1, i.e.

c = 1

∴ y = x + 1

I=12ydx=12(x+1)dx

=[x22+x]12=432

= 2.5

Definite Integrals Question 2:

The value of 0π2sin9xdxis ____.

Note: correct upto 2 decimal places.

Answer (Detailed Solution Below) 0.40 - 0.41

Definite Integrals Question 2 Detailed Solution

Concept:

0π2sinnxdx=0π2cosnxdx

If n is even, (n1)(n3)(n5)n(n2)(n4)(n6)..×π2

If n is odd, (n1)(n3)(n5)n(n2)(n4)(n6)×1                                            
Here n = 7, means n is odd,

Therefore, 0π2sin9xdx=(91)(93)(95)(97)9(92)(94)(96)(98)×1=0.4063

Definite Integrals Question 3:

The value of the following definite integral is ______ (round off to three decimal places)

e1(xlnx)dx

Answer (Detailed Solution Below) 2.090 - 2.104

Definite Integrals Question 3 Detailed Solution

Concept:

Use integration by parts for solving this problem.

Calculation:

1e(logx x)dx

={logx.x22}1e1e1x.x22dx 

={x2logx2}1e121exdx 

={x2logx2}1e{x24}1e 

={e2loge2log12}{e2414} 

=e22e24+14 

e2+14=2.097

Definite Integrals Question 4:

Given, i=1, the value of the definite integral, I=0π/2cosx+isinxcosxisinxdx is:

  1. 1
  2. -1
  3. i
  4. -i

Answer (Detailed Solution Below)

Option 3 : i

Definite Integrals Question 4 Detailed Solution

Concept:

Some important relations

1) eix=cosx+isinx

2) eix=cosxisinx

3) eiπ=cosπ+isinπ=1

4) 1i=i

Calculation:

I=0π2cosx + isinxcosx  isinxdx=0π0eixeixdx=0π2e2ixdx=12i[e2ix]0π2=12i[eiπ1]=12i(11)=1i=i

Definite Integrals Question 5:

The value of the following definite integral is π4π4sin3 x ex2x4dx

  1. π2
  2. π2
  3. π4
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Definite Integrals Question 5 Detailed Solution

Concept:

If f(x)  is even function then f(-x) = f(x)

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then aaf(x)dx=20af(x)dx

If f(x)  is odd function then aaf(x)dx=0

Calculation:

Let I = π4π4sin3 x ex2x4dx

Let f(x) = sin3 x ex2x4

Replaced x by -x, 

⇒ f(-x) = sin3 (x) e(x)2(x)4

As we know sin (-θ) = - sin θ

sin3 x ex2x4

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

Definite Integrals Question 6:

The improper definite integral xex2dx then

  1. The value of given integral is 1
  2. f(x) is odd function
  3. f(x) is even function
  4. The value of given integral is 0

Answer (Detailed Solution Below)

Option :

Definite Integrals Question 6 Detailed Solution

Explanation:

xex2dx

Letf(x)=xex2

f(x)=xex2=f(x)

∴ f(x) is odd function

∴ The value of given integral is zero.

Definite Integrals Question 7:

For the given improper integral 25dx(x2)12

  1. The integral is convergent
  2. The integral is divergent
  3. The value is ∞ 
  4. The value is 23

Answer (Detailed Solution Below)

Option :

Definite Integrals Question 7 Detailed Solution

Concept:

Improper Integral: If a function f on [a, b] have infinite value then it is called is improper integral

  1. Improper Integral of the First kind: abf(x)dx is said to be the improper integral of the first kind if a = - or b = ∞ or both.
  2. Improper Integral of the second kind: abf(x)dx is said to be the improper integral of the second kind if a or b is finite but f(x) is infinite for some x ∈ [a, b].


If the integration of the improper integral exists, then it is called as Converges, But if the limit of integration fails to exist, then the improper integral is said to be Diverge.

abdx(bx)p={;p11(1p)(ba)(p1);p<1          ---(1)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

abdx(xa)p={;p11(1p)(ba)(p1);p<1       ---(2)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

Calculation:

Given:

25dx(x2)12

On comparing with equation (2) the value of p < 1, therefore it is convergent.

25dx(x2)12=1((112)(52)(121))

=112×312=23

Definite Integrals Question 8:

Integral of sin5 x ⋅ cos2 x is:

  1. 25cos5x17cos7x13cos3x+C
  2. 25sin5+17sin7x+13sin3x+C
  3. 25cos5x+17cos7x+13cos3x+C
  4. 25sin5x17sin7x13sin3x+C

Answer (Detailed Solution Below)

Option 1 : 25cos5x17cos7x13cos3x+C

Definite Integrals Question 8 Detailed Solution

Analysis:

sin5(x)cos2(x)dx

This can be written as:

=sin(x)(sin2(x))2cos2(x)dx  ---(1)

Since sin2x + cos2x = 1:

sin2x = 1 - cos2x

Equation (1) can now be written as:

sin(x)(1cos2(x))2cos2(x)dx

Using trigonometric substitution, let u = cos (x)

dudx=sin(x)

dx=dusin(x)

sin(x)(1u2)2u2sin(x)du=u2(1u2)2du

=u2(1+u42u2)du

=(u2+u62u4)du

u2duu6du2u4du

=13u317u7+25u5+C

Substituting u = cos x back in the above expression, we get:

=25cos5(x)13cos3(x)17cos7(x)+C

Definite Integrals Question 9:

The value of integral 0π/2xcosxdx is

  1. 0
  2. -1
  3. π21
  4. π2+1

Answer (Detailed Solution Below)

Option 3 : π21

Definite Integrals Question 9 Detailed Solution

Concept:

We can solve this by using integration by parts,

∫u dv = uv - ∫ v du

let u = x; dv/dx = cos x

⇒ 0π/2xcosxdx=udv

⇒ [x(sinx)]0π/20π/2sinxdx

[xsinx+cosx]0π/2

(π2×1+0)(0+1)

π21

Definite Integrals Question 10:

The value of 14 + x2dx is:

  1. 12tan1(x2)+C
  2. 14tan1(x2)+C
  3. 14tan1(x4)+C
  4. 12tan1(x4)+C

Answer (Detailed Solution Below)

Option 1 : 12tan1(x2)+C

Definite Integrals Question 10 Detailed Solution

Concept:

Integration: 

1a2 + x2dx=1atan1(xa)+C.

Calculation:

The given expression is 14 + x2dx=122 + x2dx.

Using 1a2 + x2dx=1atan1(xa)+C, we get:

14 + x2dx=122 + x2dx=12tan1(x2)+C.

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