Solutions of Integral Equations MCQ Quiz in मल्याळम - Objective Question with Answer for Solutions of Integral Equations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 3, 2025

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

Latest Solutions of Integral Equations MCQ Objective Questions

Top Solutions of Integral Equations MCQ Objective Questions

Solutions of Integral Equations Question 1:

The definite integral 131xdx is evaluated using trapezoidal rule with a step size of 1. The correct answer is __________

  1. 1.165
  2. 1.235
  3. 2.455
  4. 1.125

Answer (Detailed Solution Below)

Option 1 : 1.165

Solutions of Integral Equations Question 1 Detailed Solution

Concept:

Trapezoidal rule is given by:

abf(x)dx=h2[yo+yn+2(y1+y2+y3)]

Numberofintervals(n)=bah

where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

Given:

f(x)=131xdx

x

1

2

3

y = f(x)

1

0.5

0.33

yn

y0

y1

y2

 

I=131xdx=h2[(y0+y2)+2y1]

=12[1+0.33+2×0.5]=2.332=1.165

Solutions of Integral Equations Question 2:

The integral  x1x2x2dx with  x2>x1>0  is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If I is the exact value of the integral obtained analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?

  1. J > I
  2. J < I
  3. J = I
  4. Insufficient data to determine the relationship

Answer (Detailed Solution Below)

Option 1 : J > I

Solutions of Integral Equations Question 2 Detailed Solution

Exact value of integration is computed by integration which follows the exact shape of graph while computing the area.

Whereas, in Trapezoidal rule, the lines joining each points are considered straight line which in not the exact variation of graph all the time

Also we know that approximate value calculated by trapezoidal rule is always greater than the exact value calculated by integration

∴ J > I

Where, J = approximate value, I = Exact value

Solutions of Integral Equations Question 3:

Evaluate the integral I=121x2.dx with 3 intervals using simpson’s 3/8th rule. Keep the accuracy upto 4 decimal places.

Answer (Detailed Solution Below) 0.5020 - 0.5022

Solutions of Integral Equations Question 3 Detailed Solution

Concept:

Simpson 3/8th rule is stated as:

abydx=3h8[(y0+yn)+2(y3+y6+)+3(y1+y2+y4)]

Where:

h=ban and 'n' must be multiple at ‘3’.

Calculation:

Given:

I=121x2dx and n = 3

 h=213=13

F2 S.B 2.6.20 Pallavi D11

I

1

4/3

5/3

2

I=1x2

1

0.5625

0.36

0.25

 

y(0)

y1

y2

y3

Using the formula of Simpson's 3/8 rule, we can write:

121x2dx=38×13[(y0+y3)+3(y1+y2)]

=18[(1+0.25)+3(0.5625+0.36)]

121x2dx=0.5021

Solutions of Integral Equations Question 4:

Using the trapezoidal rule, and dividing the interval of integration into three equal subintervals, the definite integral 1+1|x|dx is _____

Answer (Detailed Solution Below) 1.10 - 1.12

Solutions of Integral Equations Question 4 Detailed Solution

Concept:

x

-1

13

13

1

Y = |x|

1

13

13

1


N = 3

h=xnxpn=1(1)3=23

Then trapezoidal rule 11|x|dx=h2[ynyo]+2(y1+y2..+yn1)]

=232[(1+1)+2(13+13)]=109=1.1111

Solutions of Integral Equations Question 5:

For the data,

x:f(x):0512213347

the value of 042f(x)dx will be:

  1. 24
  2. 21
  3. 42
  4. 12

Answer (Detailed Solution Below)

Option 1 : 24

Solutions of Integral Equations Question 5 Detailed Solution

Concept:

Trapezoidal rule is given by:

abf(x)dx=h2[yo+yn+2(y1+y2+y3)]

Numberofintervals(n)=bah

where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

From question

x

0

1

2

3

4

f(x)

5

2

1

3

7

2f(x) 10 4 2 6 14

 

I=042.f(x).dx

From the formula:

04f(x)dx=h2[(y0+y4)+2(y1+y2+y3)]

=12[(5+7)+2(2+1+3)]

∴ I=242=12

Here we calculated for f(x), but asked for 2 f(x) 

So the answer comes out to be 2 × 12 = 24

Solutions of Integral Equations Question 6:

The integral 01(5x3+4x2+3x+2)dx

is estimated numerically using three alternative methods namely the rectangular, trapezoidal, and Simpson’s rules with a common step size. In this context, which one of the following statements is TRUE?

  1. Simpson’s rule as well as the rectangular rule of estimation will give NON-zero error.
  2. Simpson’s rule, rectangular rule as well as the trapezoidal rule of estimation will give NON-zero error.
  3. Only the rectangular rule of estimation will give zero error.
  4. Only Simpson’s rule of estimation will give zero error.

Answer (Detailed Solution Below)

Option 4 : Only Simpson’s rule of estimation will give zero error.

Solutions of Integral Equations Question 6 Detailed Solution

Concept:

General Quadrature Formula (G.Q.F):-

I=abf(x)dx=x0xnf(x)dx

=h[ny0+n22Δy0+(n33n22)Δ2y02!]

Where, h = Step size

n = Number of strips

1) Trapezoidal rule: Taking n = 1 strip at a time and neglecting second and higher-order differences in G.Q.F

F1 Neel Madhu 21.04.20 D1

x0x1f(x)dx=h[1.y0+(12)Δy0+Neglect]

=h[y0+12(y1y0)]

=h2(y0+y1)

Again,

x1x2f(x)dx=h[1.y1+12Δy1+Neglect]

=h[y1+(12)(y2y1)]

=h2(y1+y2)

Similarly, x2x3f(x)dx=h2(y2+y3) 

xn1xnf(x)dx=h2(yn1+yn)

I=abf(x)dx=h2[y0+yn+2(y1+y2+y3+yn1)]

I=h2[y0+yn+2(y1+y2+y3+yn1)]

2) Simpson’s 1/3rd Rule: If we take n = 2 strip at a time and neglect 3rd and the higher-order difference in G.Q.F

F1 Neel Madhu 21.04.20 D2

I=h3[y0+yn+4(y1+y3+y5)+2(y2+y4)]

3) Simpson’s 3/8th Rule: If we take n = 3 strips at a time and neglecting 4th and higher-order difference in G.Q.F.

I=abf(x)dx=x0x3f(x)dx+x3xbf(x)dxxn3xnf(x)dx

I=38h[y0+yn+3(y1+y2+y4+y5)+2(y3+y6+y9)]

F1 Neel Madhu 21.04.20 D3

4) Rectangle Rule

In the Rectangle rule, we approximate f|a,b| using a single interpolation point ‘a’. Our polynomial interpolant will thus be a constant polynomial p(t) = f(a), as shown in figure  and we can calculate its area IR using:

IR = f(a) ⋅ (b - a)

F1 Neel Madhu 21.04.20 D13

Thus,

 1) Trapezoidal Rule gives the exact result for a polynomial of degree 1 because we have neglected 2nd order difference in G.Q.F while the result exceeds from exact value for higher degree polynomials.

2) Simpson’s 1/3rd Rule gives the exact result for a polynomial of degree 2, while the result exceeds from exact value for higher degree polynomials.

3) Simpson’s 3/8th Rule gives the exact result for a cubic polynomial.

4) Rectangle Rule gives the exact result for a constant function.

Solutions of Integral Equations Question 7:

Match the CORRECT pairs:

Numerical

Integration Scheme

 Order of Fitting

Polynomial

P. Simpson’s 3/8 Rule  1. First
Q. Trapezoidal Rule  2. Second
R. Simpson’s 1/3 Rule 3. Third

  1. P-2; Q-1; R-3
  2. P-3; Q-2; R-1
  3. P-1; Q-2; R-3
  4. P-3; Q-1; R-2

Answer (Detailed Solution Below)

Option 4 : P-3; Q-1; R-2

Solutions of Integral Equations Question 7 Detailed Solution

Explanation:

Numberofintervals=bah

where,

b is the upper limit, a is the lower limit, h is the step size

According to the trapezoidal rule

abf(x)dx=h2[yo+yn+2(y1+y2+y3)]

It fits for 1-degree polynomial.

According to Simpson's 1/3 rule

abf(x)dx=h3[(yo+yn)+2(y2+y4+y6+)]+4[y1+y3+y5+]

It fits for 2-degree polynomial.

According to Simpson's 3/8 rule

abydx=3h8[(y0+yn)+3(y1+y2+y4+y5..)+2(y3+y6+y9..)]

It fits for 3-degree polynomial.

Solutions of Integral Equations Question 8:

The value of  02dxx+1, using trapezoidal rule with n = 4, is

  1. 1.1176
  2. 1.1167
  3. 1.119
  4. 1.1012

Answer (Detailed Solution Below)

Option 1 : 1.1176

Solutions of Integral Equations Question 8 Detailed Solution

Concept:

Numberofintervals=bah

where,

b = upper limit, a = lower limit, h = step size.

According to the trapezoidal rule

abf(x)dx=h2[yo+yn+2(y1+y2+y3)]

Calculation:

Given:

n = 4, a = 0, b = 2,  h=2040.5

y = f(x) = 1x+1

x x0 x0.5 x1 x1.5 x2
y = f(x) y0 = 1 y1 = 0.67 y2 = 0.5 y3 = 0.4 y4 = 0.33

I = 02dxx+1 = 0.52[(y0+y4)+2(y1+y2+y3)]

I = 0.25[(1 + 0.33) + 2(0.67 + 0.5 + 0.4)]

I = 0.25 [1.33 + 3.14]

I = 0.25 × 4.47

I = 1.1175

Solutions of Integral Equations Question 9:

The numerical value of the definite integral 01exdx using trapezoidal rule with function evaluations at points x = 0, 0.5 and 1 is ______ (round off to 3 decimal places)

Answer (Detailed Solution Below) 0.640 - 0.650

Solutions of Integral Equations Question 9 Detailed Solution

Concept:

Trapezoidal rule is given by:

abf(x)dx=h2[yo+yn+2(y1+y2+y3)]

Numberofintervals(n)=bah

Where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

Given:

x 0 0.5 1
f( x ) = e-x 1 0.6065 0.3678

By Trapezoidal rule:

01exdx=h2[y0+yn+2y1]

01exdx=0.52[1+0.3678+(2×0.6065)]=0.645

Solutions of Integral Equations Question 10:

Value of 45.2 ln x dx using Simpson's one-third rule with interval size 0.3 is

  1. 1.60
  2. 1.51
  3. 1.06
  4. 1.83

Answer (Detailed Solution Below)

Option 4 : 1.83

Solutions of Integral Equations Question 10 Detailed Solution

Concept:

Simpson's one-third rule:

I=h3[y0+yn+4(y1+y3+y5)+2(y2+y4)]

where n = number of intervals, interval size, h=b  an

Calculation:

Given:

f(x) = 45.2 ln x dx, h = 0.3

b = 5.2, a = 4

n=b  ah=5.2  40.3 = 4

x

4

4.3

4.6

4.9

5.2

y

ln 4

ln 4.3

ln 4.6

ln 4.9

ln 5.2

 

y0

y1

y2

y3

y4


I=h3[y0+y4+4(y1+y3)+2(y2)]

I=0.33[(ln4+ln5.2)+4(ln4.3+ln4.9)+2(ln4.6)]

∴ I = 1.8278 ≈ 1.83

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