Math / Calculus

QuestionRiemannSums24: Problem 2 Previous Problem Problem List Next Problem (1 point) The value of the limit $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{4}{n} \sqrt{2+\frac{4 i}{n}}$ is equal to the area below the graph of a function $f(x)$ on an interval $[A, B]$ . Find $f, A$ , and $B$ . (Do not evaluate the limit.) $f(x)=$ $\square$ $A=$ $\square$ (use $A=0$ ) $B=$ $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? We need to find the function $f(x)$ , and the interval $[A, B]$ that corresponds to the given Riemann sum limit. Watch out! Don't get overwhelmed by the limit notation; focus on recognizing the parts of a Riemann sum.
Remember, we don't need to evaluate the limit itself!

STEP 2

1. Deconstruct the Riemann Sum
2. Identify the Function
3. Determine the Interval

STEP 3

Remember, a Riemann sum looks like this: $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^) \cdot \Delta x$ where $\Delta x = \frac{B - A}{n}$ and $x_i^ = A + i \cdot \Delta x$ .
This represents the area under a curve $f(x)$ between $x = A$ and $x = B$ .

STEP 4

Let's look at our given limit: $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{4}{n} \sqrt{2+\frac{4i}{n}}$ We can see that $\frac{4}{n}$ plays the role of $\Delta x$ , and $\sqrt{2 + \frac{4i}{n}}$ plays the role of $f(x_i^*)$ .

STEP 5

Since $\Delta x = \frac{4}{n}$ , and $x_i^* = A + i \cdot \Delta x$ , we have $x_i^* = A + i \cdot \frac{4}{n}$ .

STEP 6

Inside the square root, we have $2 + \frac{4i}{n}$ .
If we let $A = 2$ , then $x_i^* = 2 + \frac{4i}{n}$ , which perfectly matches the expression inside the square root!

STEP 7

Since $f(x_i^) = \sqrt{2 + \frac{4i}{n}}$ and $x_i^ = 2 + \frac{4i}{n}$ , we can see that $f(x) = \sqrt{x}$ .
Awesome!

STEP 8

We know $\Delta x = \frac{B - A}{n}$ and $\Delta x = \frac{4}{n}$ .
Since we've already found that $A = 2$ , we can write $\frac{4}{n} = \frac{B - 2}{n}$ .
Multiplying both sides by $n$ gives $4 = B - 2$ , so $B = 4 + 2 = 6$ .

STEP 9

Our interval is $[A, B] = [2, 6]$ , which makes sense with our Riemann sum.

STEP 10

$f(x) = \sqrt{x}$ $A = 2$ $B = 6$

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Studdy Solution

STEP 1

What is this asking? We need to find the function $f(x)$ , and the interval $[A, B]$ that corresponds to the given Riemann sum limit. Watch out! Don't get overwhelmed by the limit notation; focus on recognizing the parts of a Riemann sum.
Remember, we don't need to evaluate the limit itself!

STEP 2

1. Deconstruct the Riemann Sum
2. Identify the Function
3. Determine the Interval

STEP 3

Remember, a Riemann sum looks like this: $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^) \cdot \Delta x$ where $\Delta x = \frac{B - A}{n}$ and $x_i^ = A + i \cdot \Delta x$ .
This represents the area under a curve $f(x)$ between $x = A$ and $x = B$ .

STEP 4

Let's look at our given limit: $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{4}{n} \sqrt{2+\frac{4i}{n}}$ We can see that $\frac{4}{n}$ plays the role of $\Delta x$ , and $\sqrt{2 + \frac{4i}{n}}$ plays the role of $f(x_i^*)$ .

STEP 5

Since $\Delta x = \frac{4}{n}$ , and $x_i^* = A + i \cdot \Delta x$ , we have $x_i^* = A + i \cdot \frac{4}{n}$ .

STEP 6

Inside the square root, we have $2 + \frac{4i}{n}$ .
If we let $A = 2$ , then $x_i^* = 2 + \frac{4i}{n}$ , which perfectly matches the expression inside the square root!

STEP 7

Since $f(x_i^) = \sqrt{2 + \frac{4i}{n}}$ and $x_i^ = 2 + \frac{4i}{n}$ , we can see that $f(x) = \sqrt{x}$ .
Awesome!

STEP 8

We know $\Delta x = \frac{B - A}{n}$ and $\Delta x = \frac{4}{n}$ .
Since we've already found that $A = 2$ , we can write $\frac{4}{n} = \frac{B - 2}{n}$ .
Multiplying both sides by $n$ gives $4 = B - 2$ , so $B = 4 + 2 = 6$ .

STEP 9

Our interval is $[A, B] = [2, 6]$ , which makes sense with our Riemann sum.

STEP 10

$f(x) = \sqrt{x}$ $A = 2$ $B = 6$

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Studdy Solution

STEP 1

STEP 2

1. Deconstruct the Riemann Sum2. Identify the Function3. Determine the Interval

STEP 3

STEP 4

STEP 5

Since Δx=4n\Delta x = \frac{4}{n}Δx=n4​, and xi∗=A+i⋅Δxx_i^* = A + i \cdot \Delta xxi∗​=A+i⋅Δx, we have xi∗=A+i⋅4nx_i^* = A + i \cdot \frac{4}{n}xi∗​=A+i⋅n4​.

STEP 6

Inside the square root, we have 2+4in2 + \frac{4i}{n}2+n4i​.If we let A=2A = 2A=2, then xi∗=2+4inx_i^* = 2 + \frac{4i}{n}xi∗​=2+n4i​, which perfectly matches the expression inside the square root!

STEP 7

Since f(xi∗)=2+4inf(x_i^*) = \sqrt{2 + \frac{4i}{n}}f(xi∗​)=2+n4i​​ and xi∗=2+4inx_i^* = 2 + \frac{4i}{n}xi∗​=2+n4i​, we can see that f(x)=xf(x) = \sqrt{x}f(x)=x​.Awesome!

STEP 8

STEP 9

Our interval is [A,B]=[2,6][A, B] = [2, 6][A,B]=[2,6], which makes sense with our Riemann sum.

STEP 10

f(x)=xf(x) = \sqrt{x}f(x)=x​ A=2A = 2A=2 B=6B = 6B=6

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1. Deconstruct the Riemann Sum
2. Identify the Function
3. Determine the Interval

Since $\Delta x = \frac{4}{n}$ , and $x_i^* = A + i \cdot \Delta x$ , we have $x_i^* = A + i \cdot \frac{4}{n}$ .

Inside the square root, we have $2 + \frac{4i}{n}$ .
If we let $A = 2$ , then $x_i^* = 2 + \frac{4i}{n}$ , which perfectly matches the expression inside the square root!

Since $f(x_i^) = \sqrt{2 + \frac{4i}{n}}$ and $x_i^ = 2 + \frac{4i}{n}$ , we can see that $f(x) = \sqrt{x}$ .
Awesome!

Our interval is $[A, B] = [2, 6]$ , which makes sense with our Riemann sum.

$f(x) = \sqrt{x}$ $A = 2$ $B = 6$