Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.2 Representing Functions - 1.2 Exercises: 42

Answer

(a) $$A(2)=4.$$ (b) $$A(6) =16.$$ (c) $$A(x)=\Bigg\{_{\frac{x^2}{2}-x+4,\quad x>2.}^{-\frac{x^2}{2}+3x,\quad x\leq2;}$$

Work Step by Step

(a) In this case, the region will be a trapezoid bounded by the $y$ axis, the $t$ axis the graph of the function and the vertical line $t=2$. This line will intercept the graph of the function at the point $(2,f(2))=(2,|2-2|+1)=(2,1)$. This trapezoid has the bases of length of $3$ and $1$ and the height of $2$ so the area is $$A(2)=\frac{3+1}{2}\cdot2=4.$$ (b) In this case we will split the region into the trapezium determined by $0\leq t\leq2$ which has the area $A(2)$ and the trapezium bounded by $t$ axis, the graph of the function and vertical lines $t=2$ and $t=6$. It has the bases of $1$ and $f(6)=|6-2|+1=5$ while its height is equal to $6-2=4$. The required area is $$A(6) = A(2) + \frac{1+5}{2}\cdot4=4+12=16.$$ (c) For $x\leq2$ the region bounded by the $y$ and $t$ axes, the graph of the function and the vertical line $t=x$ is a trapezoid with the bases of $3$ and $f(x)=|x-2|+1=2-x+1=3-x$ and the heght equal to $x$ so the area is $$A(x)=\frac{3+3-x}{2}\cdot x=-\frac{x^2}{2}+3x.$$ If $x>2$ then we have to add to the area $A(2)$ the area of the trapezoid bounded by $t$ axis, the graph of the function and the vertical lines $t=2$ and $t=x$. This trapezoid has the bases of $1$ and $f(x)=|x-2|+1=x-2+1=x-1$ and the height of $x-2$ so the required area is $$A(x)=A(2)+\frac{1+x-1}{2}\cdot(x-2)=\frac{x^2}{2}-x+4.$$ The final answer is $$A(x)=\Bigg\{_{\frac{x^2}{2}-x+4,\quad x>2.}^{-\frac{x^2}{2}+3x,\quad x\leq2;}$$
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