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## Homework Statement

Find f for:

f double prime (x) = 2x^3 + 3x^2 - 4x + 5, f(0) = 2, f(1) = 0

## Homework Equations

Most general antiderivative: F(x) + C

## The Attempt at a Solution

= F(x) + C

= f prime (x) = [(2x^4)/4] + [(3x^3) / 3] - [(4x^2)/2] + 5x + Cx

= (1/2)(x^4) + x^3 - 2x^2 + 5x + Cx

f(x) = (1/2)[(x^5)/5] + (x^4)/4 - (2x^3)/3 + (5x^2)/2 + Cx + D

The half times the x^5/5 should be further broken down, but how?

Now I'm not sure how to find Cx and D? Is the following correct?

f(0) = 0 + D = 2, therefore D = 2 If this is correct, why leave out Cx? Why was it not the other way around?

Therefore, f(x) = (1/2)[(x^5)/5] + (x^4)/4 - (2x^3)/3 + (5x^2)/2 + Cx + 2

Therefore, f(1) = (1/2)[(1^5)/5] + (1^4)/4 - (2(1)^3)/3 + (5(1)^2)/2 + Cx + 2 = 0

Therefore, 1/2(1/5) + 1/4 - 2/3 + 5/2 + 2 + Cx = 0

Therefore, Cx = 0 - 1/2(-1/5) - 1/4 + 2/3 - 5/2 - 2 Am I going correct here?