Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 29

Answer

$\frac{27}{4}$
1579346009

Work Step by Step

Step 1. Given the function $f(x)=x^3-3x^2$, we have $f'(x)=3x^2-6x$. Step 2. Let $f'(x)=0$; we have $3x(x-2)=0$ and $x=0,2$ as critical points. Step 3. Check $f''(x)=6x-6=6(x-1)$ and $f''(0)=-6\lt0$; thus the function has a local maximum at $x=0$. Step 4. Check $f''(2)=6\gt0$; thus the function has a local minimum at $x=2$. Step 5. As $f(x)=x^2(x-3)$, the zeros are $x=0$ and $x=3$ and the enclosed region between the function and the x-axis is in the interval of $[0,3]$, as shown in the figure. Step 6. The area of the enclosed region can be found as $A=-\int_0^3(x^3-3x^2)dx=-(\frac{1}{4}x^4-x^3)|_0^3=-(\frac{1}{4}(3)^4-(3)^3)=\frac{27}{4}$.
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