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Chapter 13: Problem 2

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$

Short Answer

Expert verified
The region of integration is a rectangle in the xy-plane with vertices at the points (0,0), (1,0), (0,3), and (1,3). The value of the double integral over this region is 18.

Step by step solution

01

Sketch the region of integration

The region of integration is defined by the limits of the double integral, \(0 \leq y \leq 1\) and \(0 \leq x \leq 3\). The graph of this region is a rectangle in the xy-plane with vertices at points (0,0), (1,0), (0,3), and (1,3).
02

Set up the integral

The integrand function is \(2x + 6y\). The order of integration is \( dy dx\). Therefore, the double integral is written as \[\int_{0}^{3} \int_{0}^{1}(2x + 6y) dy dx.\]
03

Evaluate the inner integral

The inner integral is \[\int_{0}^{1}(2x + 6y) dy.\]The variable of integration is y, while x is treated as a constant. Integrating gives \(2xy + 3y^2\). Then, evaluate this at y=0 and y=1, subtract the two results to obtain \(2x + 3 - 0 = 2x + 3\).
04

Evaluate the outer integral

The outer integral is \[\int_{0}^{3}(2x + 3) dx.\]This yields \(x^2 + 3x\), evaluated between 0 and 3. Subtracting the two results gives \(9 + 9 - 0 = 18\).

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Most popular questions from this chapter

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