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

1 Answer
Feb 14, 2016

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

Explanation:

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