What is #f(x) = int (x^2-2x)(e^x-3x) dx# if #f(1 ) = 4 #?
1 Answer
Feb 19, 2017
Explanation:
Note that:
#d/(dx) (x^2 e^x) = color(white)(XX) x^2 e^x + 2x e^x#
#d/(dx) (-4x e^x) = color(white)(XXX) - 4x e^x - 4e^x#
#d/(dx) (4e^x) = color(white)(XXXXXXXXXXX)4e^x#
So:
#d/(dx) (x^2-4x+4)e^x = (x^2-2x)e^x#
Hence:
#f(x) = int (x^2-2x)(e^x-3x) dx#
#color(white)(f(x)) = int (x^2-2x)e^x-3x^3+6x^2 dx#
#color(white)(f(x)) = (x^2-4x+4)e^x-3/4x^4+2x^3+C#
So:
#4 = f(1) = (1-4+4)e^1-3/4+2+C = e+5/4+C#
Hence:
#C = 4-(e+5/4) = 11/4-e#
and:
#f(x) = (x^2-4x+4)e^x-3/4x^4+2x^3+11/4-e#
