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