What is #f(x) = int e^(4x-1)-e^(3x-2)+e^x dx# if #f(2) = 3 #?

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Answer 1 · 2018-02-12T10:08:10

#f(x)=(e^(4x-1)-e^7)/4+(e^(3x-2)-e^4)/3+e^x-e^2+3#

We can write #f(x)=inte^(4x-1)-e^(3x-2)+e^xdx# as

#f(x)=inte^(4x-1)dx-inte^(3x-2)dx+inte^xdx#

Now considering #u=4x-1#, we have #du=4dx# and

#inte^(4x-1)dx=inte^u(du)/4=e^u/4=e^(4x-1)/4#

and similarly using #v=3x-2#, we can get #inte^(3x-2)dx=inte^v(dv)/3=e^v/3=e^(3x-2)/3#

and hence #f(x)=e^(4x-1)/4+e^(3x-2)/3+e^x+c#

As #f(2)=3#, we have #e^(4xx2-1)/4+e^(3xx2-2)/3+e^2+c=3#

or #e^7/4+e^4/3+e^2+c=3#

Hence #c=3-e^7/4-e^4/3-e^2# and

#f(x)=e^(4x-1)/4+e^(3x-2)/3+e^x+3-e^7/4-e^4/3-e^2#

or #f(x)=(e^(4x-1)-e^7)/4+(e^(3x-2)-e^4)/3+e^x-e^2+3#