Definite Integral Definition

[jetter2]

I am completely lost..can someone help me? I'm not sure how to approach this with the defining formula, but using the Fundamental Therom I can check my work..any pointers?

Instructions are to Evaluate the Integral with the Definition Formula using right hand rule...HALP ME!!

 

[galactus]

I suppose they mean a Riemann Sum.

You are adding up the area of an infinite number of rectangles using the right side of them as f(x). The width of each rectangle is $\displaystyle {\Delta}x$.

This is analogous to the dx in your 'regular' integral.

$\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{3-(-1)}{n}=\frac{4}{n}$

The right hand method is $\displaystyle x_{k}=a+k{\Delta}x\Rightarrow -1+\frac{4k}{n}$

$\displaystyle \left[\left(-1+\frac{4k}{n}\right)^{2}+9\left(-1+\frac{4k}{n}\right)-2\right]\left(\frac{4}{n}\right)$

Expanding and taking the sums, gives:

$\displaystyle \frac{64}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{112}{n^{2}}\sum_{k=1}^{n}k-\frac{40}{n}\sum_{k=1}^{n}1$

Note the k and k^2?. Here we sub in the closed form for the sum of consecutive integers and sum of squares.

They are $\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}, \;\ \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}, \;\ \sum_{k=1}^{n}1=n$

Sub those in.

$\displaystyle \frac{64}{n^{3}}\cdot \frac{n(n+1)(2n+1)}{6}+\frac{112}{n^{2}}\cdot \frac{n(n+1)}{2}-\frac{40}{n}\cdot n$


The algebra is a little messy, but not too bad. It whittles down and will be entirely in terms of n. Then, take the limit as $\displaystyle n\to {\infty}$.

The result is the solution to your definite integral.

 

[jetter2]

Everything makes sense, but can you clarify how you expanded tot he Sigma notation?

 

[galactus]

It's just algebra. I used the summation identities where I saw k terms.

As I showed, replace a k^2 with n(n+1)(2n+1)/6. Replace a k with n(n+1)/2.

Then, it is in terms of n alone.