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

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Answer 1 · 2016-01-12T07:38:37

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

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 2 · 2016-01-12T07:43:14

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

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#