What is f(x) = int (x+3)^2-2x dxf(x)=∫(x+3)2−2xdx if f(1) = 0 f(1)=0?

1 Answer
Øko
Dec 19, 2017

f(x)=1/3(x+3)^3-x^2-61/3f(x)=13(x+3)3−x2−613

Explanation:

f(x)=int(x+3)^2-2xdxf(x)=∫(x+3)2−2xdx

Use inth(x)+g(x)dx=inth(x)dx+intg(x)dx∫h(x)+g(x)dx=∫h(x)dx+∫g(x)dx
f(x)=int(x+3)^2dx-int2xdxf(x)=∫(x+3)2dx−∫2xdx
f(x)=1/3(x+3)^3-x^2+Cf(x)=13(x+3)3−x2+C

Evaluate the constant of integration
0=1/3(1+3)^3-1^2+C0=13(1+3)3−12+C
0=1/3(4)^3-1+C0=13(4)3−1+C
0=64/3-3/3+C0=643−33+C
0=61/3+C0=613+C
C=-61/3C=−613

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