Thread: Integrate 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

Originally Posted by norivea

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

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.

Thanks for that quick response! How come I got this answer though:

Hello, norivea!

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

Originally Posted by norivea

Thanks for that quick response! How come I got this answer though:

Your answer is 4.4646 mine is 4.67
That is close enough!
You did good.
---
Note Soroban used Rectangular rule which is completelty innacturate unless you are using many. Your and mine method using trapezoids is much more accurate.

The actual value is 4.62

Originally Posted by norivea

Thanks for that quick response! How come I got this answer though:

Because ImperfectHacker evaluated f(1.8) incorrectly. Your solution is correct.

RonL

Originally Posted by Soroban

Hello, norivea!


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

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

Software gives answer of 4.53
Which matches my answer.

Originally Posted by ThePerfectHacker

Software gives answer of 4.53
Which matches my answer.

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

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.