# Integration Problem

1. Feb 14, 2008

### JenniferBlanco

1. The problem statement, all variables and given/known data

I don't know how to make the integral sign so I am took a picture of the question.

2. Relevant equations

Equations are located in the above img.

3. The attempt at a solution

I first integrated the first derivative and got -

(ax^3)/3 + (bx^2)/2

plugged in 1 and got -

6=a + b

Then I got the second derivative --> 2ax +b and plugged in f''(1) and got: 18 = 2a +b

Using both the equations, I figured out that:
a=12
b=-6

how do I find the contant c to get f(x)?

-Jennie

2. Feb 14, 2008

### EnumaElish

By using (iii)?

I am assuming you meant f(x) = (ax^3)/3 + (bx^2)/2 + c.

3. Feb 14, 2008

### JenniferBlanco

I tried using (iii) and this is what I got -

f(x) = 4x^3 -3x^2 + c

then I plugged in 2 and 1 and got -

(32-12+c) - (4-3+c) = 18

but then c will cancel out.

I am sure I am doing this wrong. please guide me

4. Feb 14, 2008

### EnumaElish

What do you get when you integrate 4x^3 -3x^2 + c without limits?

5. Feb 14, 2008

### JenniferBlanco

x^4 - x^3 + cx

6. Feb 14, 2008

### EnumaElish

Shouldn't it be x^4-x^3+cx?

7. Feb 14, 2008

### JenniferBlanco

yes, i edited it before you replied.

8. Feb 14, 2008

### EnumaElish

What do you get for x=1? For x = 2?

9. Feb 14, 2008

### JenniferBlanco

when x=1 then I got c
when x=2 then I got 8+2c

10. Feb 14, 2008

### EnumaElish

Do you think you can use (iii)?

11. Feb 14, 2008

### JenniferBlanco

Thank you!

so,

(8 +2c) - c = 18
and c = 10

that gives me--> f(x)= 12x^2 - 6x + 10

12. Feb 14, 2008

### EnumaElish

Do you mean f(x) = 4x^3 -3x^2 + 10?

13. Feb 14, 2008

### JenniferBlanco

aaah yes. I was writing it down and was wondering how 6=18.

Sorry for the error and Thanks for the help. :)