hi
@Kainui
@sleepyjess @johnweldon1993 @Preetha
Oh, I totally need to learn this! This is the one thing my teacher didn't teach us in Calculus this year.
+ the AP test is this upcoming week zzz.
im not in calc yet. sorry.
Try reading the article I sent you and check if you can make any sense of it, because I certainly can't right now.
https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/defintsoldirectory/DefIntSol2.html#SOLUTION 10 Refer to this possibly?
Solution #10
Sampling point in your problem would be ck = 2k/n
I don't understand where the (2/3) came from in that explanation.
Oookaay, this is kinda hard to explain :P
So i'm going to try and take a intuitive explanation rather than a rigorous way to prove everything...
Hmmm, if you want to find out properly then look up the "definition of riemann integral"
I know the Riemann integral, can you explain where the 2/3 came from?
is it the delta x?
The definition of a definite integral is: \[ \int_a^b f(x) \ dx = \lim_{n\to\infty} \sum_{k=1}^n f(x_k) \Delta x = \lim_{n\to\infty} \sum_{k=1}^n f(x_k) \frac{b - a}{n} \] Try to identify the parts of your problem with the definition.
\[\int\limits_{0}^{2}\], but can you assume that? Could it possible be 4 over 2?
@andrewyates f(xk) = 5 + 2k/n
Plug in k = 1 to find the first x coordinate and then add 2 to find the bounds... I think
what would n be though?
n is the number of rectangles you're using to approximate the area under the curve. As n approaches infinity, the number of rectangles become infinite and the width of each approaches 0 so the approximation becomes exact. Here's an image: https://upload.wikimedia.org/wikipedia/commons/2/2a/Riemann_sum_convergence.png
@Hero please help savior
\[ \int_5^7(x+5)^{10} \ dx \]
Okay, I believe that's the answer.
Hey, I had the (x+5)^10 right from before xD
Nice
Until I deleted it, :P
Want me to explain? @shubham1
Can you quickly explain how you got the limits as 7 and 5?
I know they have to have a delta value of 2.
Actually, the bounds are from 0 to 2. My mistake.
\( k\frac{2}{n} \) represents the x value in the sum, so plugging in k = 1 (first x value) gives us \(\frac{2}{n}\) and as \(n\to\infty\) it becomes 0.
Add 2, and we get the bounds \(x \in [0, 2]\)
Oh wow, I had the whole equation from the start without even knowing it, haha lright thanks!
Okay, cool!
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