What are the antiderivatives of y=\frac{\ x^2-47x+33}{10} and y=-.03(x-6)^3+x?

In summary, the antiderivative of the first equation is (1/10)[(x^3/3)-(47x^2/2)+33x] and for the second equation, it is (-3/100)[(x-6)^4]+(x^2/2). A possible method for solving the second equation is using the substitution u=x-6.
  • #1
Ry122
565
2
I need help with finding the antiderivative of
1. [tex]y=\frac{\ x^2-47x+33}{10}[/tex]
and
2. [tex]y=-.03(x-6)^3+x[/tex]
My attempts:
1. [tex]y=.1[(x^3/3)-(47x^2/2)+33x)][/tex]
2. [tex]y=-.03x-.18^3+x[/tex]
[tex]y=(\frac{\-.03x^2}{2}-.18x)^3+(x^2/2)[/tex]
 
Last edited:
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  • #2
Oh I guess you edited it since I last saw it. The first integral is correct now. For the second just try a substitution.
 
Last edited:
  • #3
Come on guys there are easy ones!

1. [tex]\int \frac{\ x^2-47x+33}{10} dx = \frac{1}{10} \int x^2 - 47x + 33 dx = \frac{1}{10} ( \int x^2 dx - \int 47x dx + \int 33 dx)[/tex]

Please tell me you know the reverse power rule?

Edit: Sorry I just saw your attempt. It is correct.

2. Your attempt at the 2nd one doesn't seem correct.

[tex]\int \frac{-3}{100}(x-6)^3 + x dx = \frac{-3}{100} \int (x-6)^3 dx + \int x dx[/tex].

If you really need it, for that first integral do substitution u= x-6
 

1. What is the definition of an antiderivative?

An antiderivative is the inverse operation of differentiation, and it is a function that, when differentiated, will give the original function back.

2. What is the process for finding the antiderivative of a function?

The process for finding the antiderivative involves using a set of rules, such as the power rule, product rule, quotient rule, and chain rule, to manipulate the function until it is in a form that can be easily integrated. It is important to also include the constant of integration in the final answer.

3. Can all functions have an antiderivative?

No, not all functions have an antiderivative. Functions that do not have an antiderivative are known as non-integrable or non-elementary functions. Examples of these functions include e^x/x and sin(x^2).

4. How can the antiderivative be used in real-world applications?

The antiderivative is used in many real-world applications, including physics, engineering, and economics. It can be used to find the displacement, velocity, and acceleration of an object, as well as the area under a curve and the total cost or profit in a business.

5. Are there any shortcuts or tricks for finding the antiderivative?

Yes, there are some shortcuts or tricks for finding the antiderivative, such as using integration by parts, substitution, or trigonometric identities. However, these methods may not work for all functions, and it is important to have a strong understanding of the basic rules and techniques before attempting to use these shortcuts.

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