Q. 17 Earlier in this section, we show... [FREE SOLUTION]

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Chapter 13: Q. 17 (page 1004)

Earlier in this section, we showed that we could use Fubini鈥檚 theorem to evaluate the integral $鈭� {鈭�}_{R} x^{2} y d A$ and we showed that ${鈭�}_{2}^{5} {鈭�}_{1}^{3} x^{2} y d x d y = 91$ Now evaluate the double integral by evaluating the iterated integral ${鈭�}_{1}^{3} {鈭�}_{2}^{5} x^{2} y d y d x$ .

Short Answer

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The solution is,

${鈭�}_{1}^{3} {鈭�}_{2}^{5} x^{2} y d y d x = 91$ .

Step by step solution

Step 1. Given Information.

The iterated integral is ${鈭�}_{1}^{3} {鈭�}_{2}^{5} x^{2} y d y d x$ .

Step 2. Evaluating the integral.

We integrate with respect to $x$ .

${鈭�}_{1}^{3} {鈭�}_{2}^{5} x^{2} y d y d x = {鈭�}_{1}^{3} ({鈭�}_{2}^{5} x^{2} y d y) d x = {鈭�}_{1}^{3} ({鈭�}_{2}^{5} x^{2} (y d y)) d x = {鈭�}_{1}^{3} {[x^{2} \frac{y^{1 + 1}}{1 + 1}]}_{y = 2}^{y = 5} d x = {鈭�}_{1}^{3} {[\frac{1}{2} x^{2} y^{2}]}_{y = 2}^{y = 5} d x = {鈭�}_{1}^{3} (\frac{1}{2} x^{2} 5^{2} - \frac{1}{2} x^{2} 2^{2}) d x = {鈭�}_{1}^{3} (\frac{25}{2} x^{2} - 2 x^{2}) d x = {鈭�}_{1}^{3} (\frac{21}{2} x^{2}) d x = \frac{21}{2} {鈭�}_{1}^{3} x^{2} d x = \frac{21}{2} {[\frac{x^{2 + 1}}{2 + 1}]}_{x = 1}^{x = 3} = \frac{21}{2} {[\frac{x^{3}}{3}]}_{x = 1}^{x = 3} = \frac{21}{2} [\frac{3^{3}}{3} - \frac{1^{3}}{3}] = \frac{21}{2} [\frac{27}{3} - \frac{1}{3}] = \frac{21}{2} 脳 \frac{26}{3} = 7 脳 13 = 91$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Evaluating the integral.

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Earlier in this section, we showed that we could use Fubini鈥檚 theorem to evaluate the integral 鈭�鈭�Rx2ydAand we showed that 鈭�25鈭�13x2ydxdy=91Now evaluate the double integral by evaluating the iterated integral鈭�13鈭�25x2ydydx.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Evaluating the integral.

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Discrete Mathematics

Pure Maths

Geometry

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Probability and Statistics

Study anywhere. Anytime. Across all devices.