Problem Given the position function x(t) = t4 - 8t2, find the distance that the particle travels at t = 0 to t = 4.
Answer Key
Solution
$\displaystyle s = \int_0^4 \sqrt{1 + (4t^3 - 16t)^2} \, dt$
$s = 160.9$ ← Answer: [ A ]
Solution by Simpson's One-third Rule:
Simpson's One-third Rule may be written as $s = \dfrac{\Delta}{3} \Big[ f(t)_1 + 2\Sigma f(t)_{odd} + 4 \Sigma f(t)_{even} + f(t)_n \Big]$
Hence, $s = \dfrac{0.4}{3}(1193.15) = 159.09$ ← answer
Follow @iMATHalino
MATHalino