# Thread: Area under a curve?

1. ## Area under a curve?

Use limits to find the area between each curve and the x axis for the given interval.

1. y=(x^2+6x) from x=0 to x=4

I found delta x being 4/n and xi=4i/n

Then I plug it in

(4i/n)^2+6(4i/n) times (4/n)

I am stuck here because I see I am supposed to add but I am slightly confused in getting like term if anyone could work this part out be helpful.

(16i/n^2)+24i/n times 4/n (This is where I am stuck)

2. ## Re: Area under a curve?

Originally Posted by homeylova223
Use limits to find the area between each curve and the x axis for the given interval.

1. y=(x^2+6x) from x=0 to x=4

I found delta x being 4/n and xi=4i/n

Then I plug it in

(4i/n)^2+6(4i/n) times (4/n)

I am stuck here because I see I am supposed to add but I am slightly confused in getting like term if anyone could work this part out be helpful.

(16i/n^2)+24i/n times 4/n (This is where I am stuck)

We divide $\displaystyle $0,4$$ to $\displaystyle n$ equal parts.

$\displaystyle x_0=0,x_1=\frac{4}{n}, x_2=\frac{2}{n}\cdot4,..., x_n=4$

$\displaystyle =\mathfrak{L}(f)=\sum_{i=1}^{n}\frac{4}{n}f(x_{i-1})=$

$\displaystyle =\sum_{i=1}^{n}\frac{4}{n}(x_{i-1}^2+6x_{i-1})=$

$\displaystyle =\sum_{i=1}^{n}\frac{4}{n}((\frac{(i-1)4}{n})^2+6\frac{(i-1)4}{n})=$

$\displaystyle =\sum_{i=1}^{n}\frac{4}{n}(\frac{(i-1)4}{n})^2+\sum_{i=1}^{n}\frac{4}{n}6\frac{(i-1)4}{n}=$

$\displaystyle =\frac{4^3}{n^3}(0^2+1^2+2^2+...+(n-1)^2)+\frac{6\cdot 4^2}{n^2}(0+1+2+...+(n-1))=$

$\displaystyle =\frac{4^3}{n^3}(\frac{(n-1)n(2n-1)}{6})+\frac{6\cdot 4^2}{n^2}\frac{n(n-1)}{2}$

$\displaystyle =\mathfrak{U}(f)=\sum_{i=1}^{n}\frac{4}{n}f(x_{i}) =$

$\displaystyle =\sum_{i=1}^{n}\frac{4}{n}(x_{i}^2+6x_{i})=$

$\displaystyle =\sum_{i=1}^{n} \frac{4}{n} (( \frac{4i}{n})^2+6 \frac{4i}{n} )=$

$\displaystyle =\sum_{i=1}^{n}\frac{4}{n}(\frac{4i}{n})^2+\sum_{i =1}^{n}\frac{4}{n}6\frac{(i)4}{n}$

$\displaystyle =\frac{4^3}{n^3}(1^2+2^2+3^2+...+n^2)+\frac{6\cdot 4^2}{n^2}(1+2+...+n)=$

$\displaystyle =\frac{4^3}{n^3}(\frac{(n+1)n(2n+1)}{6})+\frac{6\c dot 4^2}{n^2}\frac{n(n+1)}{2}$

$\displaystyle \mathfrak{L}(f) \leq \int_{0}^{4}(x^2+6x)dx \leq \mathfrak{U}(f)$

When $\displaystyle n\to\infty$ we will get:

$\displaystyle \lim_{n\to\infty}\mathfrak{L}(f)=$

$\displaystyle =\lim_{n\to\infty}(\frac{4^3}{n^3}(\frac{(n-1)n(2n-1)}{6})+\frac{6\cdot 4^2}{n^2}\frac{n(n-1)}{2})=$

$\displaystyle =\frac{4^3\cdot 2}{6}+\frac{6\cdot 4^2}{2}$

$\displaystyle \lim_{n\to\infty}\mathfrak{U}(f)=$

$\displaystyle =\lim_{n\to\infty}(\frac{4^3}{n^3}(\frac{(n+1)n(2n +1)}{6})+\frac{6\cdot 4^2}{n^2}\frac{n(n+1)}{2})=$

$\displaystyle =\frac{4^3\cdot 2}{6}+\frac{6\cdot 4^2}{2}$

Hence, applying sandwich rule:

$\displaystyle \int_{0}^{4}(x^2+6x)dx=\frac{4^3\cdot 2}{6}+\frac{6\cdot 4^2}{2}$

3. ## Re: Area under a curve?

At some point, you're likely to need these:

$\displaystyle \sum_{i=1}^{n}\;i\;=\;\frac{n(n+1)}{2}$

$\displaystyle \sum_{i=1}^{n}\;i^{2}\;=\;\frac{n(n+1)(2n+1)}{6}$

I guess it took me four minutes to code that.

4. ## Re: Area under a curve?

Originally Posted by homeylova223
Use limits to find the area between each curve and the x axis for the given interval.

1. y=(x^2+6x) from x=0 to x=4

I found delta x being 4/n and xi=4i/n

Then I plug it in

(4i/n)^2+6(4i/n) times (4/n)

I am stuck here because I see I am supposed to add but I am slightly confused in getting like term if anyone could work this part out be helpful.

(16i/n^2)+24i/n times 4/n (This is where I am stuck)
You do realise that the area you are looking for is equal to $\displaystyle \displaystyle \int_0^4{x^2 + 6x\,dx}$ don't you?
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