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kogti [31]
3 years ago
6

What is the whole process of turning the limit into a definite integral? An example would be great!

Mathematics
1 answer:
juin [17]3 years ago
8 0

Step-by-step explanation:

The first step is to find the (b − a) / n factor in the limit of the Riemann sum.  This is the Δx.

The next step is to identify the function and the argument, f(xᵢ).  The argument xᵢ looks like a + i (b − a) / n.

From there, you can write two equations:

Δx = (b − a) / n

xᵢ = a + i (b − a) / n = a + i Δx

Since you know xᵢ and Δx, you can identify a.  And when you know a, you can plug that into Δx to find b.  So now you have the limits of the definite integral.

Finally, write the integral using the limits and the function.

Here's an example.  Suppose you have the limit:

lim(n→∞) ∑ᵢ₌₁ⁿ √(1 + 3i/n) (3/n)

First, we notice that Δx = (b − a)/n = 3/n.  So b − a = 3.

Next, we identify that xᵢ = 1 + 3i/n = a + i (3/n), so a = 1.  Therefore, b = 4.

Now, we identify the function, f(xᵢ) = √xᵢ.

Finally, we plug it all into a definite integral:

∫₁⁴ √x dx

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It is decreasing, since the y-value is going down.
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<h3>Answer:  -2</h3>

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Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

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-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

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