Integrating 2nd order derivative

In summary: I might be using the wrong terminology here but I thought all that I would have to do is integrate 1.75x^4+ 8x^3 - 0.5x^2+ 11x and then insert the values back in. Is this correct?
  • #1
Maths Muppet
2
0
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.
 
Physics news on Phys.org
  • #2
Hello Maths Muppet! Welcome to PF.

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

[tex]\int^b_a g'(x) dx = g(b) - g(a)[/tex]

You can apply that to b and c quite directly.
 
  • #3
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
Maths Muppet said:
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?
 
  • #5
Maths Muppet said:
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
[tex]\int_{-1}^1 f'(x)dx= f(1)- f(-1)[/tex]
and that
[tex]\int_{-1}^1 f"(x)dx= f'(1)- f'(-1)[/tex]
 
  • #6
[itex]\int_a^b f(x)\,dx\,=\,F(b)\,-\,F(a),\,where\,F'(x)\,=\,f(x).[/itex]
FTOC is great!
 

What is the definition of a 2nd order derivative?

A 2nd order derivative is the rate of change of the slope of a function. It represents how quickly the slope of a function is changing.

What is the difference between a 1st order and 2nd order derivative?

A 1st order derivative represents the rate of change of a function, while a 2nd order derivative represents the rate of change of the slope of a function.

Why is it important to integrate a 2nd order derivative?

Integrating a 2nd order derivative can help us find the original function from which the derivative was derived. It can also help us analyze the behavior of a function and determine important characteristics such as concavity and inflection points.

What are some common methods for integrating a 2nd order derivative?

Some common methods for integrating a 2nd order derivative include substitution, integration by parts, and partial fractions. Each method may be more suitable for certain types of 2nd order derivatives or functions.

What are some real-life applications of integrating 2nd order derivatives?

Integrating 2nd order derivatives has many real-life applications in fields such as physics, engineering, and economics. It can be used to model the motion of objects, analyze the stability of systems, and optimize functions for maximum efficiency.

Similar threads

Replies
31
Views
880
Replies
6
Views
863
  • Calculus
Replies
6
Views
1K
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
403
Replies
2
Views
1K
Replies
4
Views
305
Replies
1
Views
897
Replies
3
Views
1K
Back
Top