Calculate the Riemann sum S 2 , 3 for the given integral using two choices of sample points: a) Lower-left vertex b) Midpoint of rectangle Then calculate the exact value of the double integral.

Question

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Answer 1

Question

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

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Answer 2

Question

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

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Answer 3

Question

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

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Answer 4

Question

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

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Answer 5

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Answer 6

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Answer 7

Chapter 15, Problem 1CRE

To determine

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Answer 8

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The integral: $\int_{1}^{4} \int_{2}^{6} x^{2} y d x d y$

Formulas:

$S_{m, n} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i}, y_{j}) Δ A$

Where

$\begin{array}{l} Δ A = Δ x \cdot Δ y \\ Δ x = \frac{b - a}{m} and Δ y = \frac{d - c}{n} \end{array}$

Calculations:

From the given integral, we can observe that $2 \leq x \leq 6$ and $1 \leq y \leq 4$ . Since our aim is to find $S_{2, 3}$ , we need to divide the rectangle $[2, 6] \times [1, 3]$ into $2 \times 3$ subrectangles. The length and width of each subrectangle are calculated as follows:

$Δ x = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$Δ y = \frac{4 - 1}{3} = \frac{3}{3} = 1$

Therefore, the area of each subrectangle is $Δ A = Δ x \cdot Δ y = 2 \cdot 1 = 2$ .

The subrectangles are shown in Image 1.

Image 1:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum $S_{2, 3}$ . Notice that the lower-left vertices are $(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), and (4, 3)$ and are shown in Image 2.

Image 2:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 2

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (x_{i}, y_{j}) Δ A \\ = [f (x_{1}, y_{1}) + f (x_{1}, y_{2}) + f (x_{1}, y_{3}) + f (x_{2}, y_{1}) + f (x_{2}, y_{2}) + f (x_{2}, y_{3})] Δ A \\ = [f (2, 1) + f (2, 2) + f (2, 3) + f (4, 1) + f (4, 2) + f (4, 3)] \cdot 2 \\ = [2^{2} \cdot 1 + 2^{2} \cdot 2 + 2^{2} \cdot 3 + 4^{2} \cdot 1 + 4^{2} \cdot 2 + 4^{2} \cdot 3] \cdot 2 \\ = [4 + 8 + 12 + 16 + 32 + 48] \cdot 2 \\ = 120 \cdot 2 \\ = 240 \end{array}$

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum $S_{2, 3}$ . Notice that the midpoints are $(3, 1.5), (3, 2.5), (3, 3.5), (5, 1.5), (5, 2.5), and (5,3 .5)$ and are shown in Image 3.

Image 3:

Calculus: Early Transcendentals, Chapter 15, Problem 1CRE , additional homework tip 3

Thus,

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \sum_{i = 1}^{2} \sum_{j = 1}^{3} f (\bar{x_{i}}, \bar{y_{j}}) Δ A \\ = [f (\bar{x_{1}}, \bar{y_{1}}) + f (\bar{x_{1}}, \bar{y_{2}}) + f (\bar{x_{1}}, \bar{y_{3}}) + f (\bar{x_{2}}, \bar{y_{1}}) + f (\bar{x_{2}}, \bar{y_{2}}) + f (\bar{x_{2}}, \bar{y_{3}})] Δ A \\ = [f (3, 1.5) + f (3, 2.5) + f (3, 3.5) + f (5, 1.5) + f (5, 2.5) + f (5, 3.5)] \cdot 2 \\ = [3^{2} \cdot 1.5 + 3^{2} \cdot 2.5 + 3^{2} \cdot 3.5 + 5^{2} \cdot 1.5 + 5^{2} \cdot 2.5 + 5^{2} \cdot 3.5] \cdot 2 \\ = [13.5 + 22.5 + 31.5 + 37.5 + 62.5 + 87.5] \cdot 2 \\ = 255 \cdot 2 \\ = 510 \end{array}$

To calculate the exact value of the integral:

$\begin{array}{l} \int_{1}^{4} \int_{2}^{6} x^{2} y d x d y \\ = \int_{1}^{4} y d y \int_{2}^{6} x^{2} d x \\ = {\frac{y^{2}}{2}]}_{1}^{4} \cdot {\frac{x^{3}}{3}]}_{2}^{6} \\ = [\frac{4^{2}}{2} - \frac{1^{2}}{2}] \cdot [\frac{6^{3}}{3} - \frac{2^{3}}{3}] \\ = [8 - \frac{1}{2}] \cdot [72 - \frac{8}{3}] \\ = \frac{15}{2} \cdot \frac{208}{3} \\ = 520 \end{array}$

Conclusion:

Thus,

(a) The Riemann sum $S_{2, 3}$ for the given double integral using lower-left vertices is 240.

(b)The Riemann sum $S_{2, 3}$ for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Answer 12

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 6PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E

Calculate the Riemann sum S 2 , 3 for the given integral using two choices of sample points: a) Lower-left vertex b) Midpoint of rectangle Then calculate the exact value of the double integral.

Calculate the Riemann sum S2,3 for the given integral using two choices of sample points: a) Lower-left vertex b) Midpoint of rectangle Then calculate the exact value of the double integral.

Answer to Problem 1CRE

Explanation of Solution

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Chapter 15 Solutions

Calculate the Riemann sum $S_{2, 3}$ for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.