What is #F(x) = int x^2+e^(4-2x) dx# if #F(0) = 1 #?

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Answer 1 · 2016-09-21T18:04:00

# F(x)=x^3/3-e^4/2(e^(-2x)+1)+1.#

#F(x)=int(x^2+e^(4-2x))dx=intx^2dx+inte^(4-2x)dx#

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

To determine #C#, we use the cond. #F(0)=1#

#rArr 0^3/3+e^(4-0)/-2+C=1#.

#rArr C=1-e^4/2#.

Sub.img, in #F(x)#, we get,

# F(x)=x^3/3-e^(4-2x)/2+1-e^4/2, i.e., x^3/3-e^4/2(e^(-2x)+1)+1.#

Enjoy Maths.!