How do you evaluate the definite integral #int s^3-3s^2+2# from #[-2,2]#?
1 Answer
Jan 14, 2017
Explanation:
Integrate each term using the
#color(blue)"power rule for integration"#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(int(ax^n)=a/(n+1)x^(n+1))color(white)(2/2)|)))#
#rArrint_(-2)^2(s^3-3s^2+2)ds#
#=[1/4s^4-s^3+2s]_(-2)^2# Evaluate for s = 2 and subtract the evaluation for s = - 2
#=(4-8+4)-(4+8-4)=0-8=-8#