Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 287: 33

Answer

$$2$$

Work Step by Step

Given $$x+y=4, \quad x-y=0, \quad y+3 x=4$$ First, we find the intersection points; we get $x=0,\ x=1,\ x= 2$ and $$ 4-x\geq 4-3x, \ \ \ 0\leq x\leq 1\\ 4-x\geq x, \ \ \ 1\leq x\leq 2$$ Then, the area is given by \begin{aligned} \text{Area}&= A_1+A_2\\ &=\int_{0}^{1} [4-x -(4-3x)]dx+ \int_{1}^{2} [(x)-(4-x)]dx\\ &=\int_{0}^{1} [2x]dx+ \int_{1}^{2} [2x-4]dx\\ &=2x^2\bigg|_{0}^{1}+(x^2-4x)\bigg|_{1}^{2}\\ &= 2 \end{aligned} We use the Mathematica program to plot the region.
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