Use the definition of the Definite Integral to evaluate the given integral. Show all steps and make sure to use the DEFINITION not the fundamental theorem of calculus! Make use of the identities.

Question

babynini · Answer

I sort of have an idea of what i'm meant to do but not sure how to implement it o.o..ai.

babynini · Answer

@Miracrown are you available? :)

babynini · Answer

$$\int\limits_{1}^{3}x^3-x^2+2x-1=\lim_{h \rightarrow \infty}\sum_{i=1}^{n}f(x ^{*}_{i})\Delta x$$

babynini · Answer

Is the first step, I think. Although that is just a formula and now we have to find delta x

ganeshie8 · Answer

$$\int\limits_{1}^{3}x^3-x^2+2x-1=\lim_{\color{red}{n} \rightarrow \infty}\sum_{i=1}^{n}f(x ^{*}_{i})\Delta x$$

babynini · Answer

Delta x = 3-1/n = 2/n

ganeshie8 · Answer

yes

babynini · Answer

and using right end points we have
a+i(Deltax) 
= 0+i(2/n) = 2i/n

ganeshie8 · Answer

good, but here a is not 0

babynini · Answer

whoops! o.o I had written it zero when I copied it.

babynini · Answer

(2i/n)+ 1
is that how I would write it?

ganeshie8 · Answer

Yes

ganeshie8 · Answer

$x_i^* = 1+(2i/n)$


$f(x_i^*) = ?$

babynini · Answer

eem i'm not sure D:

ganeshie8 · Answer

$f(x)=x^3-x^2+2x-1$

$f(x_i^*) = ?$

babynini · Answer

oh i see

ganeshie8 · Answer

simply replace $x$ by the value of $x_i^*$

babynini · Answer

Give me a moment :)

babynini · Answer

Can I even simplify the first one? the cubed one?

ganeshie8 · Answer

you may use (a+b)^3 formula

ganeshie8 · Answer

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

babynini · Answer

hm okay. I have to go now but I shall check this in the morning and try to get my answer to you :) thanks!

ganeshie8 · Answer

good luck !

ganeshie8 · Answer

after simplifying, before evaluating the limit you should get http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5En+%28%281%2Bi*2%2Fn%29%5E3-%281%2Bi*2%2Fn%29%5E2%2B2%281%2Bi*2%2Fn%29-1%29*2%2Fn