- #1
singleton
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Good evening. I'm having a little difficulty with the summation of rectangular areas when finding the area under a curve.
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
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
Last edited: