Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 695: 33

Answer

$A= \frac{24}{5}$

Work Step by Step

The curve $x$ = $t^{3}+1$ $y$ = $2t-t^{2}$ intersects the $x$-axis when $y$ = $0$, that is when $2t-t^2=t(2-t)=0$, so $t$ = $0$ and $t$ = $2$. $A$ = $\int_0^2{[y(t)-0]x'(t)}dt$ $A$ = $\int_0^2{(2t-t^{2})(3t^{2})}dt$ $A$ = $3\int_0^2{(2t^{3}-t^{4})}dt$ $A$ = $3\left[\frac{t^{4}}{2}-\frac{t^{5}}{5}\right]_0^2$ = $\frac{24}{5}$
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