How do you evaluate the definite integral #int (x+3)(x-1)dx# from [1,3]?
1 Answer
Jan 23, 2017
Explanation:
First step is to distribute the brackets.
#rArrint_1^3(x+3)(x-1)dx#
#=int_1^3(x^2+2x-3)dx# 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);n≠-1)color(white)(2/2)|)))#
#=[1/3x^3+x^2-3x]_1^3#
#color(blue)"Note: "[F(x)]_a^b=[F(b)]-[F(a)]#
#rArr(9+9-9)-(1/3+1-3)=35/3#
