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

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Answer 1 · 2016-02-14T15:02:33

#f(x)=x^4/4-x^2+e^x+15/4-e#

We will make use of the following integration rules:

#intkf(x)dx=kintf(x)dx#

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

#inte^xdx=e^x+C#

We can integrate each term separately:

#intx^3dx=(x^(3+1))/(3+1)+C=x^4/4+C#

#int-2xdx=-2intxdx=-2(x^(1+1)/(1+1))+C=(-2x^2)/2+C=-x^2+C#

#inte^xdx=e^x+C#

Thus, our function is

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

However, we can determine #C#, the constant of integration, since we know that #f(1)=3#.

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

#3=1/4-1+e+C#

#3=-3/4+e+C#

#15/4-e=C#

We can plug this back into our function:

#f(x)=x^4/4-x^2+e^x+15/4-e#