## Calculus: Early Transcendentals 8th Edition

$\int_{0}^{3}sin(x^3)~dx \approx 0.280981$
$\Delta x = \frac{b-a}{n} = \frac{3-0}{6} = 0.5$ We can find the midpoint of each subinterval: $x_1 = 0.25$ $x_2 = 0.75$ $x_3 =1.25$ $x_4 = 1.75$ $x_5 = 2.25$ $x_6 =2.75$ $\int_{0}^{3}sin(x^3)~dx \approx \sum_{i=1}^{6} f(x_i)~\Delta x$ $=\sum_{i=1}^{6} sin(x_i^3)~\Delta x$ $= [sin(0.25^3)+sin(0.75^3)+sin(1.25^3)+sin(1.75^3)+sin(2.25^3)+sin(2.75^3)](0.5)$ $= (0.561962)(0.5)$ $= 0.280981$