Integrating 2nd order derivative

[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.

 

[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.

 

[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?

 

[rock.freak667]

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?

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?

 

[HallsofIvy]

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?

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)

 

[jhooper3581]

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