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

2 Answers
Loweena Gonasegaran
Apr 10, 2018

#7/3#

Explanation:

#int(x-3)^2-3x+4 dx#
#= (x-3)^3/3-(3x^2)/2+4x+C#
#= (2-3)^3/3-(3(2)^2)/2+4(2)+C = 4#
#= -1/3-6+8+C = 4#
#= 5/3+C = 4#
#C = 7/3#

Jim G.
Apr 10, 2018

#f(x)=1/3x^3-9/2x^2+13x-20/3#

Explanation:

#"expand "(x-3)^2" and simplify"#

#f(x)=int(x^2-6x+9-3x+4)dx#

#color(white)(f(x))=int(x^2-9x+13)dx#

#"integrate each term using the "color(blue)"power rule"#

#•color(white)(x)int(ax^n)dx=a/(n+1)x^(n+1);n!=-1#

#rArrf(x)=1/3x^3-9/2x^2+13x+c#

#"where c is the constant of variation"#

#"to find c use the condition "(f(2)=4#

#rArr4=8/3-18+26+crArrc=-20/3#

#rArrf(x)=1/3x^3-9/2x^2+13x-20/3#

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1433 views around the world
You can reuse this answer
Creative Commons License