# Integrating 2nd order derivative

1. Aug 22, 2009

### Maths Muppet

The question is

If f(x) = 7x^3 + 8x^2 - x + 11, evaluate :

a, Integral +1 - -1 f(x) dx
b, Integral +1 - -1 f'(x) dx
c, Integral +1 - -1 f''(x) dx

For a, Just integrate each individual and then input the figures which gave me

1.75x^4 + (8x^3)/3 - 0.5x^2 + 11x

Which when I input the figures gives me 27 1/3.

It is b, which I am unsure about. Do I intergrate 1.75x^4 + (8x^3)/3 - 0.5x^2 + 11x and then put the values in?

Some guidence would be most appreciated, thank you.

2. Aug 22, 2009

### Gib Z

Hello Maths Muppet! Welcome to PF.

Are you familiar with the fundamental theorem of Calculus? The form of it that is useful here is:

$$\int^b_a g'(x) dx = g(b) - g(a)$$

You can apply that to b and c quite directly.

3. Aug 24, 2009

### Maths Muppet

I think your reply has just confused me a little bit. I might be using the wrong termonology but I thought all that I would have to is integrate 1.75x^4 + 8x^3 - 0.5x^2 + 11x and then insert the values back in. Is this correct?

4. Aug 24, 2009

### rock.freak667

f'(x) means the derivative of f(x) with respect to x. Integration is the reverse of differentiation. So for example, if f(x)=x2, then f'(x)=2x. So ∫2x dx=x2+C.

See now why Gib Z said you can directly work out the integral?

5. Aug 25, 2009

### HallsofIvy

1.75x^4+ 8x^3- 0.5 x^2+ 11x is the integral of your original function and does not have to be integrated again for problem (a).

Gib_z's point is that
$$\int_{-1}^1 f'(x)dx= f(1)- f(-1)$$
and that
$$\int_{-1}^1 f"(x)dx= f'(1)- f'(-1)$$

6. Aug 28, 2009

### jhooper3581

$\int_a^b f(x)\,dx\,=\,F(b)\,-\,F(a),\,where\,F'(x)\,=\,f(x).$
FTOC is great!!!