Please tell me if I am doing the summation of rectangular areas wrongly.(adsbygoogle = window.adsbygoogle || []).push({});

Using summation of rectangles, find the area enclosed between the curve y = 3x^2 and the x-axis from x = 1 to x = 4.

Now, before I answer the way it asks, I want to use antidifferentiation first to see what I should expect.

A'(x) = F'(x) = 3x^2

Therefore F(x) = x^3

Area = F(4) - F(1)

= 64 - 1

= 63 square units, correct?

Now, using the summation of rectangles I do this:

I start by finding the width of the rectangles: (4-1)/n = 3/n

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

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

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

[tex]= 81/n^3\sum_{k=1}^n\ k^2[/tex]

[tex]= 81/n^3 * n(n+1)(2n+1)/6[/tex]

[tex]= 81/6 * n(n+1)(2n+1)/n^3[/tex]

[tex]= 27/2 * (n+1)(2n+1)/n^2[/tex]

Then I do the limit and find 27/2 * (n+1)/n * (2n+1)/n

= 27/2 * (1 + 1/n)(2 + 1/n)

As n approaches infinity, 1/n = 0

27/2 * (1 + 0)(2 + 0)

Thus I end up with 27 square units.

Two very different answers. I'm not sure what I'm doing wrong with the sums, as far as my book is concerned I'm following all "legal" rules. I've read and re-read this a few times and I don't notice any glaring mistake (on my sore eyes).

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Simple Riemann Sums question with answers not lining up

**Physics Forums | Science Articles, Homework Help, Discussion**