Good evening. I'm having a little difficulty with the summation of rectangular areas when finding the area under a curve.(adsbygoogle = window.adsbygoogle || []).push({});

Question:

Using summation of rectangles, find the area enclosed between the curve y = x^2 + 2x and the x-axis from x=0 to x=3.

Well, I start by dividing the interval (from x=0 to x=3) by n equal parts to find the width of each rectangular area.

=3/n

Then I begin using sigma notation (I'm new at this)

Sum of rectangular areas

[tex]= \sum_{k=1}^n\ f(x) * (3/n)[/tex]

[tex]= \sum_{k=1}^n\ [(k * 3/n)^2 + 2(k * 3/n)] * (3/n)[/tex]

[tex]= \sum_{k=1}^n\ [k^2 * (3/n)^2 + 2 * k * (3/n)] * (3/n)[/tex]

[tex]= \sum_{k=1}^n\ [k^2 * (3/n) + 2k] * (3/n)^2[/tex]

[tex]= 9/n^2\sum_{k=1}^n\ [k^2 * (3/n) + 2k)][/tex]

Now my main problem is that I'm trying to isolate the k^2 so that I can write out the summation formula for it and then go to limits and discover the area under the curve.

eg

[tex]\sum_{k=1}^n\ k^2[/tex]

would become

n(n+1)(2n + 1) / 6

and I could go straight to the limits

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# Homework Help: Summation of rectangular areas (calculus) problem.

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