Question

CAN SOMEONE HELP WITH THIS QUESTION?

Answer 1

49

Explanation:

`\int\limits^(12)_0 {-0.002t^(2) +0.7t} \, dt \\=(-0.002\frac{{12}^(3) }{3} )+(0.7\frac{{12}^(2) }{2} )\\-(-0.002\frac{{0}^(3) }{3} )-(0.7\frac{{0}^(2) }{2} )\\\\=(-0.002\frac{{12}^(3) }{3} )+(0.7\frac{{12}^(2) }{2} )-0-0\\\\=49.248\\`