Finding the are under the curve y=25-x^2 using infinite limits

[mathstudent1498]

I need to find the area under the curve of y=25-x^2 in the interval [1,4]. I need to use the limit of n approaches infinity, and the formula is basically this.

\(\displaystyle \displaystyle \int_a^b\, f(x)\, dx\, =\, \lim_{n\, \rightarrow\, \infty}\, \sum_{i\, =\, 1}^n\, f(x_i)\, \Delta\, x\)

I keep getting 74, when I should be getting 54.

Can someone help me out here? I think its a basic calculation error, but I can't find it, and I keep getting 74 instead of 54 each time I try the problem, even though the answer is clearly 54. I appreciate any help on this problem! Thanks!

 

[Deleted member 4993]

I need to find the area under the curve of y=25-x^2 in the interval [1,4]. I need to use the limit of n approaches infinity, and the formula is basically this. I keep getting 74, when I should be getting 54. Can someone help me out here? I think its a basic calculation error, but I can't find it, and I keep getting 74 instead of 54 each time I try the problem, even though the answer is clearly 54. I appreciate any help on this problem! Thanks!

Please show us - how did you get 74!!

 

[pka]

I need to find the area under the curve of y=25-x^2 in the interval [1,4]. I need to use the limit of n approaches infinity, and the formula is basically this. I keep getting 74, when I should be getting 54. Can someone help me out here? I think its a basic calculation error, but I can't find it, and I keep getting 74 instead of 54 each time I try the problem, even though the answer is clearly 54. I appreciate any help on this problem! Thanks!

Here is the limit notation for the right-hand Riemann Sum:
\(\displaystyle {\displaystyle\lim _{n \to \infty }}\sum\limits_{k = 1}^n {\left[ {25 - {{\left(1+ {\frac{{3k}}{n}} \right)}^2}} \right]\left( {\frac{3}{n}} \right)} \)

 

[mathstudent1498]

How I got 74

lim
n->infinite (summation of n, and I=1)((25-(i/n)^2)*((3)/n)
=(75/n-3i^2/n^3)
=(75n^2-3i^2)/n^3
I then take out the n^3 as common, and use summation formulas to get the following
75n^3-3(2n^3+n^2+2n^2+n)/6
=75n^3(+(n^3-2n^3-n^2-2n^2-n)/2)
=150n^3-2n^3-n^2-2n^2-n/2
I then multiply by the n^3 I factored out above
150n^3-2n^3-n^2-2n^2-n/2n^3
The limit at infinity is 74, while it should be 54.
So that's how I got 74.

 

[mathstudent1498]

Please show us - how did you get 74!!

 

[mathstudent1498]

Please show us - how did you get 74!!

I can't upload the other picture, so I'll write it out.
75n^3-3((2n^3+n^2+2n^2+n)/6
=75n^3-(2n^3-n^2-2n^2-n)/2
=(150n^3-2n^3-n^2-2n^2-n)/2
I now multiply the n^3 I factored out earlier.
150n^3-2n^3-n^2-2n^2-n/2n^3
This comes out to the lim 148n^3/2n^3= 74
n->infinity

 

[mathstudent1498]

Thank you for your help everyone!

Thank you for your help everyone! I figured out I wasn't using the correct formula to find x, a+i *change in x was necessary. If I hadn't been shown the Riemann sums I would have had a hard time figuring this out.

 

[pka]

lim
n->infinite (summation of n, and I=1)((25-(i/n)^2)*((3)/n) That is wrong.
=(75/n-3i^2/n^3)
=(75n^2-3i^2)/n^3
I then take out the n^3 as common, and use summation formulas to get the following
75n^3-3(2n^3+n^2+2n^2+n)/6
=75n^3(+(n^3-2n^3-n^2-2n^2-n)/2)
=150n^3-2n^3-n^2-2n^2-n/2
I then multiply by the n^3 I factored out above
150n^3-2n^3-n^2-2n^2-n/2n^3
The limit at infinity is 74, while it should be 54.
So that's how I got 74.

You are integrating over the interval \(\displaystyle [1,4]\) so you must subdivide that interval into \(\displaystyle n\) equal parts:
\(\displaystyle \left[1+\frac{(k-1)3}{n},1+\frac{(k)3}{n}\right],~k=1,\cdots n\).
Before you try any more of this you need a sit-down with a live instructor to help you sort it all out.