What is #F(x) = int 3x^2-e^(2-x) dx# if #F(0) = 1 #?

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Answer 1 · 2017-05-15T17:35:00

#F(x)=x^3+e^(2-x)+1-e^2, or, #

#F(x)=x^3+e^2(e^-x -1)+1.#

#F(x)=int3x^2-e^(2-x)dx.#

#=3intx^2dx-inte^2*e^-xdx,#

#=3(x^3/3)-e^2inte^-xdx#

#F(x)=x^3-e^2*(e^-x/-1)+C=x^3+e^(2-x)+C.#

But, #F(0)=1 rArr 0^3+e^(2-0)+C=1.#

# rArr C=1-e^2.#

Therefore, #F(x)=x^3+e^(2-x)+1-e^2, or, #

#F(x)=x^3+e^2(e^-x -1)+1.#