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

2 Answers
Loweena Gonasegaran
Apr 10, 2018

7/373

Explanation:

int(x-3)^2-3x+4 dx∫(x−3)2−3x+4dx
= (x-3)^3/3-(3x^2)/2+4x+C=(x−3)33−3x22+4x+C
= (2-3)^3/3-(3(2)^2)/2+4(2)+C = 4=(2−3)33−3(2)22+4(2)+C=4
= -1/3-6+8+C = 4=−13−6+8+C=4
= 5/3+C = 4=53+C=4
C = 7/3C=73

Jim G.
Apr 10, 2018

f(x)=1/3x^3-9/2x^2+13x-20/3f(x)=13x3−92x2+13x−203

Explanation:

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

f(x)=int(x^2-6x+9-3x+4)dxf(x)=∫(x2−6x+9−3x+4)dx

color(white)(f(x))=int(x^2-9x+13)dxf(x)=∫(x2−9x+13)dx

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

•color(white)(x)int(ax^n)dx=a/(n+1)x^(n+1);n!=-1∙x∫(axn)dx=an+1xn+1;n≠−1

rArrf(x)=1/3x^3-9/2x^2+13x+c⇒f(x)=13x3−92x2+13x+c

"where c is the constant of variation"where c is the constant of variation

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

rArr4=8/3-18+26+crArrc=-20/3⇒4=83−18+26+c⇒c=−203

rArrf(x)=1/3x^3-9/2x^2+13x-20/3⇒f(x)=13x3−92x2+13x−203

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