Finding the area by dividing into sub intervals.

I have to find the area of the following by dividing it up into nine sub intervals.

$\displaystyle W(x)=-x^4/324+x^3/9-25x^2/18+7x, 0<x<18$

the width of each interval is 2, so nine sub intervals ought to be x=2,4,6,8,9,10,12,14,16,18......i multiplied the width by the f(2,4,6..etc) and the answer came out to be approx 178.96, however when i calculate the integral on tech the answer is 183.6....have i done something wrong?

Re: Finding the area by dividing into sub intervals.

Quote:

Originally Posted by

**johnsy123** I have to find the area of the following by dividing it up into subintevals.

3) Your backyard pool is kidney shaped and its width can be modelled as a function of its length (*x*) using the rule

[IMG]file:///C:/Users/Nich/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]

(a) Use a numerical method to calculate the total area of the pool by finding the area under this graph, for example divide the interval up into nine subintervals and add up the areas.

You need to post images through a website, not from your computer.

Re: Finding the area by dividing into sub intervals.

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Re: Finding the area by dividing into sub intervals.

Quote:

Originally Posted by

**johnsy123** I have to find the area of the following by dividing it up into nine sub intervals.

$\displaystyle W(x)=-x^4/324+x^3/9-25x^2/18+7x, 0<x<18$

the width of each interval is 2, so nine sub intervals ought to be x=2,4,6,8,9,10,12,14,16,18......i multiplied the width by the f(2,4,6..etc) and the answer came out to be approx 178.96, however when i calculate the integral on tech the answer is 183.6....have i done something wrong?

What you have done appears to be alright, but right-hand-interval integration won't give you a very accurate answer. You could improve it by averaging it with the left-hand-interval integral, or the midpoint rule, or the trapezoidal rule, or Simpson's Rule.