Unit – 4

Numerical Integration

Q1) What is numerical integration?

A1)

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In case of function of single variable, the process is called quadrature.

Q2) What is Newton cotes formula?

A2)

Suppose where y takes the values

And let the integration interval (a,b) is divided into n equal sub-intervals, each of width h = b – a /n, so that,

x0 = a, x1 = x0 + h, x2 = x0 + 2h, . . . , xn = x0 + nh = b.

I =

The above formula is known as Newton’s cotes formula.

This is also known as general quadrature formula.

Q3) What is trapezoidal rule?

A3)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

Q4) State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

A4)

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

X | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |

Y | 23 | 19 | 14 | 11 | 12.5 | 16 | 19 | 20 | 20 |

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Q5) Compute the value of ?

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

A5)

Here

For h=0.5, we construct the data table:

X | 0 | 0.5 | 1 |

Y | 1 | 0.8 | 0.5 |

By Trapezoidal rule

For h=0.25, we construct the data table:

X | 0 | 0.25 | 0.5 | 0.75 | 1 |

Y | 1 | 0.94117 | 0.8 | 0.64 | 0.5 |

By Trapezoidal rule

For h = 0.125, we construct the data table:

X | 0 | 0.125 | 0.25 | 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 |

Y | 1 | 0.98461 | 0.94117 | 0.87671 | 0.8 | 0.71910 | 0.64 | 0.56637 | 0.5 |

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q6) Evaluate, using trapezoidal rule with five ordinates

A6)

Here

We construct the data table:

X | 0 | |||||

Y | 0 | 0.3693161 | 1.195328 | 1.7926992 | 1.477265 | 0 |

Q7) What is Simpson’s rule?

A7)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Note: In this rule third and higher differences are neglected an so f(x) is a polynomial of degree 2.

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively

A8)

For n=4, we have

Construct the data table:

X | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |

Y=1/x | 1 | 0.66666 | 0.5 | 0.4 | 0.33333 |

By Simpson’s Rule

For n = 8, we have

X | 1 | 1.25 | 1.50 | 1.75 | 2.0 | 2.25 | 2.50 | 2.75 | 3.0 |

Y=1/x | 1 | 0.8 | 0.66666 | 0.571428 | 0.5 | 0.444444 | 0.4 | 0.3636363 | 0.333333 |

By Simpson’s Rule

Q9) Evaluate

A9)

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X | 0 | ||||||

0 | 0.50874 | 0.707106 | 0.840896 | 0.930604 | 0.98281 | 1 |

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X | 3 | 4 | 5 | 6 | 7 |

9.88751 | 22.108709 | 40.23594 | 64.503340 | 95.34959 |

By Simpson’s Rule

= 177.3853

Q11) What is Simpson’s 3/8th rule?

A11)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3 , we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Q12) Evaluate

By Simpson’s 3/8 rule.

A12)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

1.3863 | 1.4351 | 1.4816 | 1.5261 | 1.5686 | 1.6094 | 1.6487 |

By Simpson’s 3/8 rule

= 1.8278475

Q13) Evaluate .

A13)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 0.5 | 0.2 | 0.1 | 0.0588 | 0.0385 | 0.027 |

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

Q14) A river is 80 m wide. The depth ‘b’ of the river at a distance ‘a’ from one bank is given by the table-

a | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

B | 0 | 4 | 7 | 9 | 12 | 15 | 14 | 8 | 3 |

Find the approximate area of cross-section of the river using Boole’s rule.

A14)

The area of the cross-section of the river will be-

The number of sub-intervals here is 8.

Then by Boole’s rule-

= 2(10)/45 [ 7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) + 12(14) + 32(8) +7(3)]

= 708

So that the area of of the cross-section of the river is 708 square meter.

Q15) Evaluate by using Boole’s.

A15)

Take h = , so that there four sub intervals-

X | 1 | 2 | 3 | 4 | 5 |

F(x) | 1 | ½ | 1/3 | ¼ | 1/5 |

Using Boole’s method-

Q16) Compute the following integral upto three decimal places by using Romberg’s method.

A16)

We take h = 0.5, 0.25 and 0.125 successively and use the results as below

I(h) = 0.7084, I(1/2h) = 0.6970 and I(1/4 h) = 0.6941

Hence,

We get

Finally,

The table of values,

0.7084

0.6932

0.6970 0.6931

0.6931

0.6941

Unit – 4

Numerical Integration

Q1) What is numerical integration?

A1)

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In case of function of single variable, the process is called quadrature.

Q2) What is Newton cotes formula?

A2)

Suppose where y takes the values

And let the integration interval (a,b) is divided into n equal sub-intervals, each of width h = b – a /n, so that,

x0 = a, x1 = x0 + h, x2 = x0 + 2h, . . . , xn = x0 + nh = b.

I =

The above formula is known as Newton’s cotes formula.

This is also known as general quadrature formula.

Q3) What is trapezoidal rule?

A3)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

Q4) State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

A4)

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

X | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |

Y | 23 | 19 | 14 | 11 | 12.5 | 16 | 19 | 20 | 20 |

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Q5) Compute the value of ?

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

A5)

Here

For h=0.5, we construct the data table:

X | 0 | 0.5 | 1 |

Y | 1 | 0.8 | 0.5 |

By Trapezoidal rule

For h=0.25, we construct the data table:

X | 0 | 0.25 | 0.5 | 0.75 | 1 |

Y | 1 | 0.94117 | 0.8 | 0.64 | 0.5 |

By Trapezoidal rule

For h = 0.125, we construct the data table:

X | 0 | 0.125 | 0.25 | 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 |

Y | 1 | 0.98461 | 0.94117 | 0.87671 | 0.8 | 0.71910 | 0.64 | 0.56637 | 0.5 |

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q6) Evaluate, using trapezoidal rule with five ordinates

A6)

Here

We construct the data table:

X | 0 | |||||

Y | 0 | 0.3693161 | 1.195328 | 1.7926992 | 1.477265 | 0 |

Q7) What is Simpson’s rule?

A7)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Note: In this rule third and higher differences are neglected an so f(x) is a polynomial of degree 2.

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively

A8)

For n=4, we have

Construct the data table:

X | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |

Y=1/x | 1 | 0.66666 | 0.5 | 0.4 | 0.33333 |

By Simpson’s Rule

For n = 8, we have

X | 1 | 1.25 | 1.50 | 1.75 | 2.0 | 2.25 | 2.50 | 2.75 | 3.0 |

Y=1/x | 1 | 0.8 | 0.66666 | 0.571428 | 0.5 | 0.444444 | 0.4 | 0.3636363 | 0.333333 |

By Simpson’s Rule

Q9) Evaluate

A9)

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X | 0 | ||||||

0 | 0.50874 | 0.707106 | 0.840896 | 0.930604 | 0.98281 | 1 |

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X | 3 | 4 | 5 | 6 | 7 |

9.88751 | 22.108709 | 40.23594 | 64.503340 | 95.34959 |

By Simpson’s Rule

= 177.3853

Q11) What is Simpson’s 3/8th rule?

A11)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3 , we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Q12) Evaluate

By Simpson’s 3/8 rule.

A12)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

1.3863 | 1.4351 | 1.4816 | 1.5261 | 1.5686 | 1.6094 | 1.6487 |

By Simpson’s 3/8 rule

= 1.8278475

Q13) Evaluate .

A13)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 0.5 | 0.2 | 0.1 | 0.0588 | 0.0385 | 0.027 |

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

Q14) A river is 80 m wide. The depth ‘b’ of the river at a distance ‘a’ from one bank is given by the table-

a | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

B | 0 | 4 | 7 | 9 | 12 | 15 | 14 | 8 | 3 |

Find the approximate area of cross-section of the river using Boole’s rule.

A14)

The area of the cross-section of the river will be-

The number of sub-intervals here is 8.

Then by Boole’s rule-

= 2(10)/45 [ 7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) + 12(14) + 32(8) +7(3)]

= 708

So that the area of of the cross-section of the river is 708 square meter.

Q15) Evaluate by using Boole’s.

A15)

Take h = , so that there four sub intervals-

X | 1 | 2 | 3 | 4 | 5 |

F(x) | 1 | ½ | 1/3 | ¼ | 1/5 |

Using Boole’s method-

Q16) Compute the following integral upto three decimal places by using Romberg’s method.

A16)

We take h = 0.5, 0.25 and 0.125 successively and use the results as below

I(h) = 0.7084, I(1/2h) = 0.6970 and I(1/4 h) = 0.6941

Hence,

We get

Finally,

The table of values,

0.7084

0.6932

0.6970 0.6931

0.6931

0.6941

Unit – 4

Numerical Integration

Q1) What is numerical integration?

A1)

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In case of function of single variable, the process is called quadrature.

Q2) What is Newton cotes formula?

A2)

Suppose where y takes the values

And let the integration interval (a,b) is divided into n equal sub-intervals, each of width h = b – a /n, so that,

x0 = a, x1 = x0 + h, x2 = x0 + 2h, . . . , xn = x0 + nh = b.

I =

The above formula is known as Newton’s cotes formula.

This is also known as general quadrature formula.

Q3) What is trapezoidal rule?

A3)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

Q4) State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

A4)

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

X | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |

Y | 23 | 19 | 14 | 11 | 12.5 | 16 | 19 | 20 | 20 |

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Q5) Compute the value of ?

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

A5)

Here

For h=0.5, we construct the data table:

X | 0 | 0.5 | 1 |

Y | 1 | 0.8 | 0.5 |

By Trapezoidal rule

For h=0.25, we construct the data table:

X | 0 | 0.25 | 0.5 | 0.75 | 1 |

Y | 1 | 0.94117 | 0.8 | 0.64 | 0.5 |

By Trapezoidal rule

For h = 0.125, we construct the data table:

X | 0 | 0.125 | 0.25 | 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 |

Y | 1 | 0.98461 | 0.94117 | 0.87671 | 0.8 | 0.71910 | 0.64 | 0.56637 | 0.5 |

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q6) Evaluate, using trapezoidal rule with five ordinates

A6)

Here

We construct the data table:

X | 0 | |||||

Y | 0 | 0.3693161 | 1.195328 | 1.7926992 | 1.477265 | 0 |

Q7) What is Simpson’s rule?

A7)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Note: In this rule third and higher differences are neglected an so f(x) is a polynomial of degree 2.

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively

A8)

For n=4, we have

Construct the data table:

X | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |

Y=1/x | 1 | 0.66666 | 0.5 | 0.4 | 0.33333 |

By Simpson’s Rule

For n = 8, we have

X | 1 | 1.25 | 1.50 | 1.75 | 2.0 | 2.25 | 2.50 | 2.75 | 3.0 |

Y=1/x | 1 | 0.8 | 0.66666 | 0.571428 | 0.5 | 0.444444 | 0.4 | 0.3636363 | 0.333333 |

By Simpson’s Rule

Q9) Evaluate

A9)

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X | 0 | ||||||

0 | 0.50874 | 0.707106 | 0.840896 | 0.930604 | 0.98281 | 1 |

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X | 3 | 4 | 5 | 6 | 7 |

9.88751 | 22.108709 | 40.23594 | 64.503340 | 95.34959 |

By Simpson’s Rule

= 177.3853

Q11) What is Simpson’s 3/8th rule?

A11)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3 , we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Q12) Evaluate

By Simpson’s 3/8 rule.

A12)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

1.3863 | 1.4351 | 1.4816 | 1.5261 | 1.5686 | 1.6094 | 1.6487 |

By Simpson’s 3/8 rule

= 1.8278475

Q13) Evaluate .

A13)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 0.5 | 0.2 | 0.1 | 0.0588 | 0.0385 | 0.027 |

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

Q14) A river is 80 m wide. The depth ‘b’ of the river at a distance ‘a’ from one bank is given by the table-

a | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

B | 0 | 4 | 7 | 9 | 12 | 15 | 14 | 8 | 3 |

Find the approximate area of cross-section of the river using Boole’s rule.

A14)

The area of the cross-section of the river will be-

The number of sub-intervals here is 8.

Then by Boole’s rule-

= 2(10)/45 [ 7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) + 12(14) + 32(8) +7(3)]

= 708

So that the area of of the cross-section of the river is 708 square meter.

Q15) Evaluate by using Boole’s.

A15)

Take h = , so that there four sub intervals-

X | 1 | 2 | 3 | 4 | 5 |

F(x) | 1 | ½ | 1/3 | ¼ | 1/5 |

Using Boole’s method-

Q16) Compute the following integral upto three decimal places by using Romberg’s method.

A16)

We take h = 0.5, 0.25 and 0.125 successively and use the results as below

I(h) = 0.7084, I(1/2h) = 0.6970 and I(1/4 h) = 0.6941

Hence,

We get

Finally,

The table of values,

0.7084

0.6932

0.6970 0.6931

0.6931

0.6941

Unit – 4

Unit – 4

Unit – 4

Numerical Integration

Q1) What is numerical integration?

A1)

Q2) What is Newton cotes formula?

A2)

Suppose where y takes the values

x0 = a, x1 = x0 + h, x2 = x0 + 2h, . . . , xn = x0 + nh = b.

I =

The above formula is known as Newton’s cotes formula.

This is also known as general quadrature formula.

Q3) What is trapezoidal rule?

A3)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

A4)

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

X | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |

Y | 23 | 19 | 14 | 11 | 12.5 | 16 | 19 | 20 | 20 |

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Q5) Compute the value of ?

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

A5)

Here

For h=0.5, we construct the data table:

X | 0 | 0.5 | 1 |

Y | 1 | 0.8 | 0.5 |

By Trapezoidal rule

For h=0.25, we construct the data table:

X | 0 | 0.25 | 0.5 | 0.75 | 1 |

Y | 1 | 0.94117 | 0.8 | 0.64 | 0.5 |

By Trapezoidal rule

For h = 0.125, we construct the data table:

X | 0 | 0.125 | 0.25 | 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 |

Y | 1 | 0.98461 | 0.94117 | 0.87671 | 0.8 | 0.71910 | 0.64 | 0.56637 | 0.5 |

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q6) Evaluate, using trapezoidal rule with five ordinates

A6)

Here

We construct the data table:

X | 0 | |||||

Y | 0 | 0.3693161 | 1.195328 | 1.7926992 | 1.477265 | 0 |

Q7) What is Simpson’s rule?

A7)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively

A8)

For n=4, we have

Construct the data table:

X | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |

Y=1/x | 1 | 0.66666 | 0.5 | 0.4 | 0.33333 |

By Simpson’s Rule

For n = 8, we have

X | 1 | 1.25 | 1.50 | 1.75 | 2.0 | 2.25 | 2.50 | 2.75 | 3.0 |

Y=1/x | 1 | 0.8 | 0.66666 | 0.571428 | 0.5 | 0.444444 | 0.4 | 0.3636363 | 0.333333 |

By Simpson’s Rule

Q9) Evaluate

A9)

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X | 0 | ||||||

0 | 0.50874 | 0.707106 | 0.840896 | 0.930604 | 0.98281 | 1 |

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X | 3 | 4 | 5 | 6 | 7 |

9.88751 | 22.108709 | 40.23594 | 64.503340 | 95.34959 |

By Simpson’s Rule

= 177.3853

Q11) What is Simpson’s 3/8th rule?

A11)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3 , we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Q12) Evaluate

By Simpson’s 3/8 rule.

A12)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

1.3863 | 1.4351 | 1.4816 | 1.5261 | 1.5686 | 1.6094 | 1.6487 |

By Simpson’s 3/8 rule

= 1.8278475

Q13) Evaluate .

A13)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 0.5 | 0.2 | 0.1 | 0.0588 | 0.0385 | 0.027 |

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

a | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

B | 0 | 4 | 7 | 9 | 12 | 15 | 14 | 8 | 3 |

Find the approximate area of cross-section of the river using Boole’s rule.

A14)

The area of the cross-section of the river will be-

The number of sub-intervals here is 8.

Then by Boole’s rule-

= 2(10)/45 [ 7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) + 12(14) + 32(8) +7(3)]

= 708

So that the area of of the cross-section of the river is 708 square meter.

Q15) Evaluate by using Boole’s.

A15)

Take h = , so that there four sub intervals-

X | 1 | 2 | 3 | 4 | 5 |

F(x) | 1 | ½ | 1/3 | ¼ | 1/5 |

Using Boole’s method-

Q16) Compute the following integral upto three decimal places by using Romberg’s method.

A16)

We take h = 0.5, 0.25 and 0.125 successively and use the results as below

I(h) = 0.7084, I(1/2h) = 0.6970 and I(1/4 h) = 0.6941

Hence,

We get

Finally,

The table of values,

0.7084

0.6932

0.6970 0.6931

0.6931

0.6941

Unit – 4

Unit – 4

Numerical Integration

Q1) What is numerical integration?

A1)

Q2) What is Newton cotes formula?

A2)

Suppose where y takes the values

x0 = a, x1 = x0 + h, x2 = x0 + 2h, . . . , xn = x0 + nh = b.

I =

The above formula is known as Newton’s cotes formula.

This is also known as general quadrature formula.

Q3) What is trapezoidal rule?

A3)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

A4)

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

X | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |

Y | 23 | 19 | 14 | 11 | 12.5 | 16 | 19 | 20 | 20 |

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Q5) Compute the value of ?

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

A5)

Here

For h=0.5, we construct the data table:

X | 0 | 0.5 | 1 |

Y | 1 | 0.8 | 0.5 |

By Trapezoidal rule

For h=0.25, we construct the data table:

X | 0 | 0.25 | 0.5 | 0.75 | 1 |

Y | 1 | 0.94117 | 0.8 | 0.64 | 0.5 |

By Trapezoidal rule

For h = 0.125, we construct the data table:

X | 0 | 0.125 | 0.25 | 0.375 | 0.5 | 0.625 | 0.75 | 0.875 | 1 |

Y | 1 | 0.98461 | 0.94117 | 0.87671 | 0.8 | 0.71910 | 0.64 | 0.56637 | 0.5 |

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q6) Evaluate, using trapezoidal rule with five ordinates

A6)

Here

We construct the data table:

X | 0 | |||||

Y | 0 | 0.3693161 | 1.195328 | 1.7926992 | 1.477265 | 0 |

Q7) What is Simpson’s rule?

A7)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively

A8)

For n=4, we have

Construct the data table:

X | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |

Y=1/x | 1 | 0.66666 | 0.5 | 0.4 | 0.33333 |

By Simpson’s Rule

For n = 8, we have

X | 1 | 1.25 | 1.50 | 1.75 | 2.0 | 2.25 | 2.50 | 2.75 | 3.0 |

Y=1/x | 1 | 0.8 | 0.66666 | 0.571428 | 0.5 | 0.444444 | 0.4 | 0.3636363 | 0.333333 |

By Simpson’s Rule

Q9) Evaluate

A9)

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X | 0 | ||||||

0 | 0.50874 | 0.707106 | 0.840896 | 0.930604 | 0.98281 | 1 |

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X | 3 | 4 | 5 | 6 | 7 |

9.88751 | 22.108709 | 40.23594 | 64.503340 | 95.34959 |

By Simpson’s Rule

= 177.3853

Q11) What is Simpson’s 3/8th rule?

A11)

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3 , we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Q12) Evaluate

By Simpson’s 3/8 rule.

A12)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

1.3863 | 1.4351 | 1.4816 | 1.5261 | 1.5686 | 1.6094 | 1.6487 |

By Simpson’s 3/8 rule

= 1.8278475

Q13) Evaluate .

A13)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 0.5 | 0.2 | 0.1 | 0.0588 | 0.0385 | 0.027 |

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

a | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

B | 0 | 4 | 7 | 9 | 12 | 15 | 14 | 8 | 3 |

Find the approximate area of cross-section of the river using Boole’s rule.

A14)

The area of the cross-section of the river will be-

The number of sub-intervals here is 8.

Then by Boole’s rule-

= 2(10)/45 [ 7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) + 12(14) + 32(8) +7(3)]

= 708

So that the area of of the cross-section of the river is 708 square meter.

Q15) Evaluate by using Boole’s.

A15)

Take h = , so that there four sub intervals-

X | 1 | 2 | 3 | 4 | 5 |

F(x) | 1 | ½ | 1/3 | ¼ | 1/5 |

Using Boole’s method-

Q16) Compute the following integral upto three decimal places by using Romberg’s method.

A16)

We take h = 0.5, 0.25 and 0.125 successively and use the results as below

I(h) = 0.7084, I(1/2h) = 0.6970 and I(1/4 h) = 0.6941

Hence,

We get

Finally,

The table of values,

0.7084

0.6932

0.6970 0.6931

0.6931

0.6941