What is #f(x) = int (x^2-2x)(e^x-3x) dx# if #f(1 ) = 2 #?

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Answer 1 · 2017-07-24T18:28:45

#f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e#

#f(x)=int(x^2-2x)*e^x*dx-int(3x^3-6x^2)*dx#

= #[(x^2-2x)-(2x-2)+2]*e^x-(3/4*x^4-2x^3)+C#

= #(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+C#

After imposing #f(1)=2# condition,

#(1-4+4)*e-(3/4-2)+C=2#

#C=3/4-e#

Thus, #f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e#

1) I integrated the right side of the equation.

2) I found C for #f(1)=2# condition.