What is f(x) = int x^4-x^2+5x dx if f(3)=-1 ?

1 Answer
Jan 31, 2016

f(x)=x^5/5-x^3/3+(5x^2)/2-631/10

Explanation:

Integrate each term individually using the rule:

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

Applying this, we see that

f(x)=(x^(4+1))/(4+1)-(x^(2+1))/(2+1)+(5x^(1+1))/(1+1)+C

f(x)=x^5/5-x^3/3+(5x^2)/2+C

Since we know that f(3)=-1, we can find C, the constant of integration.

f(3)=(3^5)/5-3^3/3+(5(3^2))/2+C=-1

243/5-9+45/2+C=-1

486/10-90/10+225/10+C=-10/10

C=-631/10

Thus,

f(x)=x^5/5-x^3/3+(5x^2)/2-631/10