How do you evaluate #\int _ { x } ^ { 2} ( 3t ^ { 2} + 8t ) d t#?
1 Answer
Jul 4, 2017
# int_x^2 \ (3t^2+8t) \ dt = 24 - x^3 - 4x^2#
Explanation:
# int_x^2 \ (3t^2+8t) \ dt = [(3t^3)/3 + (8t^2)/2 ]_x^2#
# " " = [t^3 + 4t^2]_x^2#
# " " = (2^3+4 * 2^2) - (x^3 + 4 * x^2)#
# " " = (8+16) - (x^3 + 4x^2)#
# " " = 24 - x^3 - 4x^2#
