Use a numerical method to find the area enclosed by the graph of y = x² sin x, the x axis and the lines x = 1 and x = 2.6

Use an interval of 0.4

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- October 9th 2006, 01:41 PMnoriveaIntegrate functions and use calculus to solve problems
Use a numerical method to find the area enclosed by the graph of y = x² sin x, the x axis and the lines x = 1 and x = 2.6

Use an interval of 0.4 - October 9th 2006, 02:29 PMThePerfectHacker
We can approximate via trapezoid rule for the function is continous.

In order to use the trapezoid rule tou need to find the number of intervals.

In this case the length is 2.6-1=1.6

Then divided by the length of each one 1.6/.4=4

Thus, a use of 4 intervals.

By the trapezoid rule we need to find,

f(1),f(1.4),f(1.8),f(2.2),f(2.6)

We note that,

f(1)=(1)^2*sin(1)=0.84

f(1.4)=(1.4)^2*sin(1.4)=1.93

f(1.8)=(1.8)^2*sin(1.8)=3.16

f(2.2)=(2.2)^2*sin(2.2)=3.91

f(2.6)=(2.6)^2*sin(2.6)=3.48

Now we double the inner terms (look at formula)

Thus,

f(1)=.84

2f(1.4)=3.86

2f(1.8)=7.32

2f(2.2)=7.82

f(2.6)=3.48

Add them together,

Result=23.32

Now multiply it by the length of each interval, (.4)

Thus,

9.33

Finally multiply by 1/2,

4.67

That is the approximate answer. - October 9th 2006, 02:58 PMnorivea
Thanks for that quick response! How come I got this answer though:

- October 9th 2006, 03:08 PMSoroban
Hello, norivea!

Quote:

Use a numerical method to**approximate**the area enclosed by:

the graph of y = x² sin x, the x axis and the lines x = 1 and x = 2.6

Use an interval of 0.4

There are many available methods.

I'll use left-endpoints.

. f(1) . = . .1²·sin(1) . . ≈ .0.8415

f(1.4) .= .1.4²·sin(1.4) .≈ .1.9315

f(1.8) .= .1.8²·sin(1.8) .≈ .3.1553

f(2.2) .= .2.2²·sin(2.2) .≈ .3.9131

f(2.4) .= .2.4²·sin(2.4) .≈ .3.8907

f(2.6) .= .2.6²·sin(2.6) .≈ .3.4845

Area .≈ .(0.8415 + 1.9315 + 3.1553 + 3.9131 + 3.8907 + 3.4845) x 0.4 .= .**6.88676**

- October 9th 2006, 03:56 PMThePerfectHacker
- October 10th 2006, 01:58 AMCaptainBlack
- October 10th 2006, 02:04 AMCaptainBlack
When using the left end point method we dont use your last point f(2.6),

also all the increments should be equal.

So your solution should look like:

. f(1) . = . .1²·sin(1) . . ≈ .0.8415

f(1.4) .= .1.4²·sin(1.4) .≈ .1.9315

f(1.8) .= .1.8²·sin(1.8) .≈ .3.1553

f(2.2) .= .2.2²·sin(2.2) .≈ .3.9131

Area .≈ .(0.8415 + 1.9315 + 3.1553 + 3.9131) x 0.4 .=3.94

RonL - October 10th 2006, 06:25 AMThePerfectHacker
Software gives answer of 4.53

Which matches my answer. - October 10th 2006, 06:43 AMCaptainBlack
The actual value of the integral is of no relevance here, we have an

excercise in numerical integration with a specified step size. The

answer for any particular method is very unlikely to agree with the

answer obtained analyticaly.

Also 4.46 is closer to 4.53 than your answer of 4.67.

RonL - October 10th 2006, 06:49 AMThePerfectHacker
I now realized what the user was complaining about.

When it wrote,

4.464646

I saw that as,

4.646464

And that was close to my answer = 4.67

But that is not.