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

Question

1259 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-20T20:17:05

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-8#
#f(x)=e^(2x)/2-e^x+x^2/2-1444#

Let's split this into:
#inte^(2x)dx-inte^xdx+intxdx#

The two rules of integrating for this are:
#inte^(ax+b)dx=e^(ax+b)/a+C#
#intx^ndx=x^(n+1)/(n+1)+C#

This gives us:
#f(x)=e^(2x)/2-e^x+x^2/2+C#

Though we need #C#, however we are told that #f(4)=0#

#0=e^(2(4))/2-e^4+4^2/2+C#

#0=e^8/2-e^4+8+C#

#C=e^4-e^8/2-8~~-1444#

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-8#
#f(x)=e^(2x)/2-e^x+x^2/2-1444#