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.

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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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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:

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition), 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

Want to see more full solutions like this?

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.