Math / Calculus

Question```latex Suppose $f(x) = \frac{x^{2}}{9}$ .
(a) The rectangles in the graph on the left illustrate a left endpoint Riemann sum for $f(x)$ on the interval $2 \leq x \leq 4$ . The value of this left endpoint Riemann sum is $\square$ and it is an $\_\_\_\_\_\_\_\_\_$ .
(b) The rectangles in the graph on the right illustrate a right endpoint Riemann sum for $f(x)$ on the interval $2 \leq x \leq 4$ . The value of this right endpoint Riemann sum is $\square$ and it is an $\_\_\_\_\_\_\_\_\_$ $\square$ the area of the region enclosed by $y = f(x)$ , the $x$ -axis, and the vertical lines $x = 2$ and $x = 4$ .

Studdy Solution

STEP 1

What is this asking? We're looking at two different ways to estimate the area under a curve using rectangles, one using left endpoints and the other using right endpoints, and we need to figure out which estimate is bigger! Watch out! Don't mix up left and right endpoints!
Also, remember that these are just estimates of the area, not the exact area.

STEP 2

1. Calculate the Left Riemann Sum
2. Calculate the Right Riemann Sum
3. Compare the Sums

STEP 3

Alright, let's break down this left Riemann sum!
We're looking at the interval from $x = 2$ to $x = 4$ , and it's split into four equal subintervals.
This means each rectangle has a width of $\frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2} = \mathbf{0.5}$ .

STEP 4

For the left Riemann sum, the height of each rectangle is determined by the function's value at the left endpoint of each subinterval.
The left endpoints are $x = 2$ , $x = 2.5$ , $x = 3$ , and $x = 3.5$ .

STEP 5

Let's calculate those heights using our function $f(x) = \frac{x^2}{9}$ : $f(2) = \frac{2^2}{9} = \frac{4}{9}$ $f(2.5) = \frac{(2.5)^2}{9} = \frac{6.25}{9}$ $f(3) = \frac{3^2}{9} = \frac{9}{9} = 1$ $f(3.5) = \frac{(3.5)^2}{9} = \frac{12.25}{9}$

STEP 6

Now, let's find the area of each rectangle by multiplying its width ( $\mathbf{0.5}$ ) by its height.
Then, we'll add them all up to get the left Riemann sum: $L = 0.5 \cdot \left( \frac{4}{9} + \frac{6.25}{9} + \frac{9}{9} + \frac{12.25}{9} \right) = 0.5 \cdot \frac{31.5}{9} = 0.5 \cdot 3.5 = \mathbf{1.75}$

STEP 7

Time for the right Riemann sum!
We still have the same interval and the same four subintervals, so the width of each rectangle is still $\mathbf{0.5}$ .

STEP 8

This time, the height of each rectangle is the function's value at the right endpoint of each subinterval.
The right endpoints are $x = 2.5$ , $x = 3$ , $x = 3.5$ , and $x = 4$ .

STEP 9

We already calculated $f(2.5)$ , $f(3)$ , and $f(3.5)$ in the previous steps.
We just need $f(4)$ : $f(4) = \frac{4^2}{9} = \frac{16}{9}$

STEP 10

Let's calculate the right Riemann sum: $R = 0.5 \cdot \left( \frac{6.25}{9} + \frac{9}{9} + \frac{12.25}{9} + \frac{16}{9} \right) = 0.5 \cdot \frac{43.5}{9} = 0.5 \cdot 4.8333... = \mathbf{2.41666...}$

STEP 11

Our left Riemann sum is $\mathbf{1.75}$ , and our right Riemann sum is $\mathbf{2.41666...}$ .
Since the function $f(x) = \frac{x^2}{9}$ is increasing on the interval $[2, 4]$ , the right Riemann sum will always be an overestimate of the actual area, and the left Riemann sum will always be an underestimate.

STEP 12

(a) The left endpoint Riemann sum is $\mathbf{1.75}$ and it is an underestimate. (b) The right endpoint Riemann sum is $\mathbf{2.41666...}$ and it is an overestimate of the area.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Calculate the Left Riemann Sum
2. Calculate the Right Riemann Sum
3. Compare the Sums

STEP 3

Alright, let's break down this left Riemann sum!
We're looking at the interval from $x = 2$ to $x = 4$ , and it's split into four equal subintervals.
This means each rectangle has a width of $\frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2} = \mathbf{0.5}$ .

STEP 4

For the left Riemann sum, the height of each rectangle is determined by the function's value at the left endpoint of each subinterval.
The left endpoints are $x = 2$ , $x = 2.5$ , $x = 3$ , and $x = 3.5$ .

STEP 5

Let's calculate those heights using our function $f(x) = \frac{x^2}{9}$ : $f(2) = \frac{2^2}{9} = \frac{4}{9}$ $f(2.5) = \frac{(2.5)^2}{9} = \frac{6.25}{9}$ $f(3) = \frac{3^2}{9} = \frac{9}{9} = 1$ $f(3.5) = \frac{(3.5)^2}{9} = \frac{12.25}{9}$

STEP 6

STEP 7

Time for the right Riemann sum!
We still have the same interval and the same four subintervals, so the width of each rectangle is still $\mathbf{0.5}$ .

STEP 8

This time, the height of each rectangle is the function's value at the right endpoint of each subinterval.
The right endpoints are $x = 2.5$ , $x = 3$ , $x = 3.5$ , and $x = 4$ .

STEP 9

We already calculated $f(2.5)$ , $f(3)$ , and $f(3.5)$ in the previous steps.
We just need $f(4)$ : $f(4) = \frac{4^2}{9} = \frac{16}{9}$

STEP 10

Let's calculate the right Riemann sum: $R = 0.5 \cdot \left( \frac{6.25}{9} + \frac{9}{9} + \frac{12.25}{9} + \frac{16}{9} \right) = 0.5 \cdot \frac{43.5}{9} = 0.5 \cdot 4.8333... = \mathbf{2.41666...}$

STEP 11

STEP 12

(a) The left endpoint Riemann sum is $\mathbf{1.75}$ and it is an underestimate. (b) The right endpoint Riemann sum is $\mathbf{2.41666...}$ and it is an overestimate of the area.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Calculate the Left Riemann Sum2. Calculate the Right Riemann Sum3. Compare the Sums

STEP 3

STEP 4

For the *left* Riemann sum, the height of each rectangle is determined by the function's value at the *left* endpoint of each subinterval.The left endpoints are x=2x = 2x=2, x=2.5x = 2.5x=2.5, x=3x = 3x=3, and x=3.5x = 3.5x=3.5.

STEP 5

STEP 6

STEP 7

Time for the *right* Riemann sum!We still have the same interval and the same **four** subintervals, so the width of each rectangle is still 0.5\mathbf{0.5}0.5.

STEP 8

This time, the height of each rectangle is the function's value at the *right* endpoint of each subinterval.The right endpoints are x=2.5x = 2.5x=2.5, x=3x = 3x=3, x=3.5x = 3.5x=3.5, and x=4x = 4x=4.

STEP 9

We already calculated f(2.5)f(2.5)f(2.5), f(3)f(3)f(3), and f(3.5)f(3.5)f(3.5) in the previous steps.We just need f(4)f(4)f(4): f(4)=429=169f(4) = \frac{4^2}{9} = \frac{16}{9}f(4)=942​=916​

STEP 10

STEP 11

STEP 12

(a) The left endpoint Riemann sum is 1.75\mathbf{1.75}1.75 and it is an **underestimate**. (b) The right endpoint Riemann sum is 2.41666...\mathbf{2.41666...}2.41666... and it is an **overestimate** of the area.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Calculate the Left Riemann Sum
2. Calculate the Right Riemann Sum
3. Compare the Sums

For the left Riemann sum, the height of each rectangle is determined by the function's value at the left endpoint of each subinterval.
The left endpoints are $x = 2$ , $x = 2.5$ , $x = 3$ , and $x = 3.5$ .

Time for the right Riemann sum!
We still have the same interval and the same four subintervals, so the width of each rectangle is still $\mathbf{0.5}$ .

This time, the height of each rectangle is the function's value at the right endpoint of each subinterval.
The right endpoints are $x = 2.5$ , $x = 3$ , $x = 3.5$ , and $x = 4$ .

We already calculated $f(2.5)$ , $f(3)$ , and $f(3.5)$ in the previous steps.
We just need $f(4)$ : $f(4) = \frac{4^2}{9} = \frac{16}{9}$

(a) The left endpoint Riemann sum is $\mathbf{1.75}$ and it is an underestimate. (b) The right endpoint Riemann sum is $\mathbf{2.41666...}$ and it is an overestimate of the area.