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

2 Answers
Jan 12, 2016

Answer:

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

Explanation:

For solving the problem first integrate the indefinite integral.

#f(x)=int quad x^3-x dx#

Use rule #int quad x^n dx = x^(n+1)/(n+1) + C#

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

We are given #f(2)=4#
We are going to find #C# using the given value.

#f(2)=(2)^4/4-(2)^2/2+C#
#4=16/4-4/2+C#
#4=4-2+C#
#4=2+C #C=2#

Therefore,

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

Answer:

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

Explanation:

Given: #f(x) = int(x^3-x) ##dx# and #f(2)=4#

#f(x)=int(x^3-x)##dx# = #x^4/4-x^2/2+C#

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

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

and #C=2#

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