Use the Definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=x^{2}+\sqrt{1+2 x}, \quad 6 \leq x \leq 8
$

$\lim _{x \rightarrow \infty} \sum_{i=1}^{8} \square$

# Question 2

The area $A$ of the region $S$ that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.

$
A=\lim _{x \rightarrow \infty} R_{n}=\lim _{x \rightarrow \infty}\left[f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\ldots+f\left(x_{n}\right) \Delta x\right]
$

Use this definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=\frac{8 f(x)}{x}, 6 \leq x \leq 11
$

$A=\lim _{x \rightarrow \infty} \sum_{i=1}^{8} \square$

## Example with answer

The area $A$ of the region $S$ that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.

$
A=\lim _{x \rightarrow \infty} R_{n}=\lim _{x \rightarrow \infty}\left[f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\ldots+f\left(x_{n}\right) \Delta x\right]
$

Use this definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=\sqrt[3]{x}, 1 \leq x \leq 12
$

$A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \quad \sqrt{\sqrt{1+\frac{11 i}{n}} \frac{11}{n}}$

The answer were not.

Question

Use the Definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=x^{2}+\sqrt{1+2 x}, \quad 6 \leq x \leq 8
$

$\lim _{x \rightarrow \infty} \sum_{i=1}^{8} \square$

# Question 2

The area $A$ of the region $S$ that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.

$
A=\lim _{x \rightarrow \infty} R_{n}=\lim _{x \rightarrow \infty}\left[f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\ldots+f\left(x_{n}\right) \Delta x\right]
$

Use this definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=\frac{8 f(x)}{x}, 6 \leq x \leq 11
$

$A=\lim _{x \rightarrow \infty} \sum_{i=1}^{8} \square$

## Example with answer

The area $A$ of the region $S$ that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.

$
A=\lim _{x \rightarrow \infty} R_{n}=\lim _{x \rightarrow \infty}\left[f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\ldots+f\left(x_{n}\right) \Delta x\right]
$

Use this definition to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$
f(x)=\sqrt[3]{x}, 1 \leq x \leq 12
$

$A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \quad \sqrt{\sqrt{1+\frac{11 i}{n}} \frac{11}{n}}$

The answer were not.

StudyXY · Accepted Answer

: Identify the function and the interval : Recall the definition of the area as a limit of the sum of areas of approximating rectangles : Apply the definition to the given function and interval : Substitute the function f(x) into the expression The expression for the area A under the graph of f(x) = x^2 + √(1 + 2x) on the interval [6, 8] as a limit is: $ A = \lim\_{n 	o \infty} \sum\_{i=1}^{n} \left[\left(6 + i \* \frac{2}{n}ight)^2 + \sqrt{1 + 2\left(6 + i \* \frac{2}{n}ight)}ight] \frac{2}{n} $ Note: This expression is not evaluated; we are only finding the expression for the area as a limit.

Q
QuestionAnatomy and Physiology

Answer

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Step 1
: Identify the function and the interval

Step 2
: Recall the definition of the area as a limit of the sum of areas of approximating rectangles

Final Answer

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