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

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Answer 1 · 2016-01-31T02:42:01

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

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#