Help with calculating area with limit definition?

Question

using the limiti definition, calculate the area of a region bounded by th efunction f(x)=2x-x^3 and the x-axis on the interval [0,1].

Answer 1

An integral always represents a sum. In most area determinations, it represents a sum of tiny little (vertical) rectangles and you sum the area of each one.

The width of these rectangles is represented by dx. Their height is represented by y = f(x) = 2x - x³

So you are summing from (i=0 to k) of:
A(approx) = (width = 1/k) * (height = 2(i/k) - (i/k)³)

or (rewriting)

A(approx) = Sum from i=0 to k of:
1/k * (2i/k - i³/k³)

Now the sum of 1 + 2 + 3 .... + k is k(k+1)/2
The sum of 1 + 8 + 27 + ... + k³ is k²(k+1)²/4

So A(approx) = 1/k * (k(k+1)/k - k²(k+1)²/(4k³))
But this has just been an approximation. The smaller the widths of our tiny little rectangles, the better the approximation is. So we write that:

A(lim) = limit as k approaches infinity of:
(k+1)/k - (k+1)²/(4k²)
which is just:
A = 1 - 1/4 = 3/4

Answer 2

I'm assuming that you have not been "exposed" to Reimann sums or derivatives.

You can try direct substitution with the given limit values which evaluates to:

2(1) - 1^3 = 1.
You can see by making a table also, which we learned prior to taking Calculus. Although if you plot the points on a graphing calculator, the area beneath the curve (the limit definition) is not exactly 1.

Answer 3

A = x^2- x^4/4 evaluated from 0 to 1
= 1- 1/4 -0 = 3/4

Answer 4

Use Riemann sums.

I use SUM to indicate sigma notation.
1) lim(w->0) SUM(n goes from 0 to 1/w) [2(nw)-(nw)^3]w
2) lim w SUM 2nw-n^3w^3
3) lim w (SUM 2nw - SUM n^3w^3)
4) lim w (2w SUM n - w^3 SUM n^3)
5) lim w (2w ((1/w)(1/w+1)/2) - w^3 ((1w)(1/w+1)/2)^2)
6) lim w [((w+1)/w) - 1/4 ((w+1)^2/w)]
7) lim w+1 - 1/4 (w+1)^2
=1-1/4
=3/4

In step 1, I wrote the Riemann sum
In step 2, I expanded the parentheses and took a w out of the Sigma notation (since the summation didn't involve w)
In step 3, I separated the summation into two parts.
In step 4, I took w out of summation again.
In step 5, I used formulas for SUM n and SUM n^3
In step 6, I simplified what I got in step 5
In step 7, I just took the limit as w goes to 0.

Sorry for the messy equations, this was the best I could do. I hope it helps!

Answer 5

delta x=1/n
x*=0+(k/n)
k

Area=lim as n->infinity (sigma from k=1 to n) of f(x*)*delta x
k

f(x*)=(2k/n)-(k^3/n^3)
k
which equals (2kn^2-k^3)/(n^3)*(1/n) is (2kn^2-k^3)/(n^4)
Now use sigma notation on the terms,

(2/n^2)sigma k -(2/n^4)sigma k^3
Which equals (2/n^2)((n^2+n)/2)-(2/n^4)( (n^2+n)/2)^2)
Solve from there and use the limit as n approaches infinity.