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

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Answer 1 · 2017-03-05T10:28:38

The answer is #=3/4x^4-2/2x^2+e^x(x-1)+13/4#

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=-1)#

We start, by calculating

#intxe^xdx#

#intuv'dx=uv-intu'vdx#

#u=x#, #=>#, #u'=1#

#v'=e^x#, #=>#, #v=e^x#

Therefore,

#intxe^xdx=xe^x-inte^xdx=xe^x-e^x#

So,

#f(x)=int(3x^3-2x+xe^x)dx#

#=3intx^3dx-2intxdx+intxe^xdx#

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

#f(1)=3/4-1+0+C=3#

#C=3+1/4=13/4#

Therefore,

#f(x)=3/4x^4-2/2x^2+e^x(x-1)+13/4#