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

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

Latest Properties of Definite Integrals MCQ Objective Questions

Top Properties of Definite Integrals MCQ Objective Questions

Properties of Definite Integrals Question 1:

Evaluate: 0π2sinx(1+cos2x)dx

  1. π4
  2. π2
  3. π 
  4. 0

Answer (Detailed Solution Below)

Option 1 : π4

Properties of Definite Integrals Question 1 Detailed Solution

Concept:

Some useful formulas are:

ddx(cosx)=sinx

1(1+x2)dx=tan1x+c

tan-11 = π4

tan-10 = 0

pqf(x)dx=qpf(x)dx

Calculation:

0π2sinx(1+cos2x)dx

Let, cosx = z

∴ - sinx dx = dz 

Change of limit
x 0 π2
z 1 0
 

Substituting the values we get,

101(1+z2)dz

011(1+z2)dz

tan1z|01

= tan-11 - tan-10

π4

Properties of Definite Integrals Question 2:

The value of 2  2(ax5+bx3+c) dx depends on the value of:

  1. b.
  2. c.
  3. a.
  4. a and b.

Answer (Detailed Solution Below)

Option 2 : c.

Properties of Definite Integrals Question 2 Detailed Solution

Concept:

For an odd function f(x): a  af(x) dx=0.

Calculation:

We observe that ax5 and bx3 are odd functions of x, 

∵ a(-x)5 = -ax5 and b(-x)3 = -bx3.

2  2ax5 dx=0 and 2  2bx3 dx=0.

The value of 2  2(ax5+bx3+c) dx depends on the value of c.

In fact, 

2  2(ax5+bx3+c) dx=2  2c dx.

= c [2 - (-2)]

= 4c

Additional Information

A function f(x) is:

  • Even, if f(-x) = f(x). And a  af(x) dx=20af(x) dx.
  • Odd, if f(-x) = -f(x). And a  af(x) dx=0.

Properties of Definite Integrals Question 3:

Evaluate, π/2π/211+etanxdx .

  1. 0
  2. 1
  3. π/2 
  4. π/4 

Answer (Detailed Solution Below)

Option 3 : π/2 

Properties of Definite Integrals Question 3 Detailed Solution

Concept:

Definite Integral properties:

abf(x)dx=abf(a+bx)dx

Trigonometry Formula:

tan ( -x ) = - tan x 

Calculation:

Let , I = π/2π/211+etanxdx                 ....(1)

Now using property, abf(x)dx=abf(a+bx)dx

I = π/2π/211+etan(x)dx

I = π/2π/211+etanxdx                    .... (ii)

Adding equation (i) and (ii), we get

2I = π/2π/2{11+etanx+11+etanx}dx  

2I = π/2π/2{11+etanx+etanx1+etanx}dx 

2I = π/2π/21+etanx1+etanxdx 

2I = π/2π/21dx 

2I = π

∴  I = π /2  

The correct option is 3. 

Properties of Definite Integrals Question 4:

If f(x) and g(x) are continuous functions satisfying f(x) = f(a – x) and g(x) + g(a – x) = 2, then what is 0af(x)g(x)dx equal to?

  1. 0ag(x)dx
  2. 0af(x)dx
  3. 20af(x)dx
  4. 0

Answer (Detailed Solution Below)

Option 2 : 0af(x)dx

Properties of Definite Integrals Question 4 Detailed Solution

Concept:

Integral Properties:

  • abf(x)dx=abf(a+bx)dx


Calculation:

Given: f(x) = f(a – x) & g(x) + g(a – x) = 2

To find: 0af(x)g(x)dx

I=0af(x)g(x)dx      …. (1)

Using integral property, abf(x)dx=abf(a+bx)dx

I=0af(ax)g(ax)dx

Using given, f(x) = f(a – x)

I=0af(x)g(ax)dx      …. (2)

Adding equation (1) & (2)

2I=0af(x)g(x)dx+0af(x)g(ax)dx

2I=0af(x){g(x)+g(ax)}dx

Using given, g(x) + g(a – x) = 2

2I=0af(x)×2dx

I=0af(x)dx

Properties of Definite Integrals Question 5:

The value of ππcot1x dx will be

  1. -π 
  2. π 
  3. 2 
  4. π2 

Answer (Detailed Solution Below)

Option 4 : π2 

Properties of Definite Integrals Question 5 Detailed Solution

Concept:

Some useful formulas are:

abf(x)dx=abf(a+bx)dx

cot1(x)=(πcot1x) dx

xndx=(xn+1)(n+1)+C; n1

Calculation:

Let, I = ππcot1x dx

Since, abf(x)dx=abf(a+bx)dx , so

 I = ππcot1(π+πx) dx

∴ I = ππcot1(x) dx

∴ I = ππ(πcot1x) dx

∴ I = πππ dx - ππcot1x dx

∴  I = πππ dx - I

∴ 2I = πx|ππ=π[π(π)]=2π2

∴ I = π2 

Properties of Definite Integrals Question 6:

What is value of π/2π/2x5sin4xdx ?

  1. 1
  2. π/2 
  3. π/4 
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Properties of Definite Integrals Question 6 Detailed Solution

Concept: 

The rules for integrating even and odd functions ,

If the function is even or odd and the interval is [-a, a], we can apply these rules:

  • When f(x) is even ⇔ aaf(x)dx=20af(x)dx
  • When f(x) is odd ⇔ aaf(x)dx=0 

Calculation: 

Let f (x) = x5 sin4 x 

f (-x) = (-x)5 sin4 (-x) = -x5 (-sin x )= -x5 sin4 x .

⇒ f(-x) = - f(x) 

So, f(x) is an odd function . 

We knoiw that, when f(x) is odd ⇔ aaf(x)dx=0 

Hence , π/2π/2x5sin4xdx = 0 .  

The correct option is 4 .

Properties of Definite Integrals Question 7:

If 0π/2logcosx dx=π2log(12) then 0π/2logsecx dx=

  1. π2log(12)
  2. 1π2log(12)
  3. 1+π2log(12)
  4. π2log2

Answer (Detailed Solution Below)

Option 4 : π2log2

Properties of Definite Integrals Question 7 Detailed Solution

Calculation:

Given: 0π/2logcosx dx=π2log(12)          .... (1)

Now,

LetI=0π/2logsecx dx=0π/2log(1cosx) dx

=0π/2log1 dx0π/2logcosx dx        (∵ log mn=logmlogn)

=00π/2logcosx dx                            (∵ log 1 = 0)

From equation (1), we get

I=π2log(12)=π2[log1log2]

π2log2

Properties of Definite Integrals Question 8:

Evaluate the integral 0π2(3sinx+4cosx)sinx+cosxdx

  1. π4
  2. 7π2
  3. 7π4
  4. 7π5

Answer (Detailed Solution Below)

Option 3 : 7π4

Properties of Definite Integrals Question 8 Detailed Solution

Concept:

Property of definite integrals: 

0af(x)dx=0af(ax)dx

Calculation:

I = 0π2(3sinx+4cosx)sinx+cosxdx         …. (1)

Using the property of definite integrals: 

0af(x)dx=0af(ax)dx

=0π2[3sin(π2x)+4cos(π2x)]sin(π2x)+cos(π2x)dx

=0π2(3cosx+4sinx)cosx+sinxdx       .... (2)

Adding equations (1) and (2), we get

2I=0π2(3sinx+4cosx)sinx+cosxdx+0π2(3cosx+4sinx)cosx+sinxdx

=0π2(7cosx+7sinx)cosx+sinxdx

=0π27(cosx+sinx)cosx+sinxdx

=0π27dx

=[7x]0π2

=7(π20)

=7π2

∴ I = 7π4

Properties of Definite Integrals Question 9:

Let f and g be continuous functions on [0, a] such that f(x) = f(a - x) and g(x) + g(a - x)=4, then 0af(x)g(x)dx is equal to 

  1. 40af(x)dx
  2. 20af(x)dx
  3. 30af(x)dx
  4. 0af(x)dx

Answer (Detailed Solution Below)

Option 2 : 20af(x)dx

Properties of Definite Integrals Question 9 Detailed Solution

Concept:

Property of definite integral:

0af(x)dx=0af(ax)dx

Calculation:

Given, f(x) = f(a - x)  __(i)

and g(x) + g(a - x) = 4  ___(ii)

 I = 0af(x)g(x)dx___(iii)

⇒ I = 0af(ax)g(ax)dx {Using property}

⇒ I = 0af(x)g(ax)dx {Using (i)}  __(iv)

Adding equation (iii) and (iv),

⇒ 2I = 0af(x)g(x)dx + 0af(x)g(ax)dx

⇒ 2I = 0af(x)[g(x)+g(ax)]dx

⇒ 2I = 0a4f(x)dx {Using (ii)}

⇒ I = 20af(x)dx

∴ The correct answer is option (2).

Properties of Definite Integrals Question 10:

The value of π/4π/4sin103x.cos101xdx is

  1. 0
  2. 2
  3. (π4)101
  4. (π4)103

Answer (Detailed Solution Below)

Option 1 : 0

Properties of Definite Integrals Question 10 Detailed Solution

Concept:

Properties of definite integral:

abf(x)dx = abf(b+ax)dx

Calculation:
I = π/4π/4sin103x.cos101xdx                 ...(i)

I = π/4π/4sin103(π4+π4x).cos101(π4+π4x)dx

I = π/4π/4sin103(x).cos101(x)dx

I = π/4π/4(sinx)103.(cosx)101dx

I = π/4π/4sin103x.cos101xdx    ...(ii)

Adding (i) and (ii), we get

2I = π/4π/4sin103x.cos101x+(sin103x.cos101x)dx

2I = 0

I = 0

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