# Integrating (4x-3)^2: Finding a Primitive Function

• Mattara
In summary, the task was to find one primitive function to (4x-3)^2. Two methods were used, one involving expanding and integrating the polynomial directly, and the other involving using the chain rule backwards. Both methods yielded different results, but both were correct with different constant values. If the problem had been to find all primitive functions, the answer would have been the same for both methods with a family of functions with different constant values.
Mattara
The task is to find one primitive function to:

$$(4x-3)^2$$

This was quite straightforward. Or so I though.

It can easily be turned into

$$16x^2 - 24x + 9$$

and then integrated to

$$\frac {16x^3}{3} - 12x^2 + 9x$$

choosing C = 0

Then I started to think. Couldn't this be integrated using the chain rule backwards (with lack of better wording)?

$$(4x-3)^2$$

then becomes according to my integration

$$\frac {(4x-3)^3}{12}$$

If I differentiate the above, I get the initial function. What I'm having a hard time understanding is that the different approaches yields different results. I plotted both of them in my graph calculator and they indeed give different results.

Any hints on what I did wrong is greatly appreciated. Thank you for your time. Have a nice day.

When you plot both of them, what exactly is the difference? Try expanding the second function, and see if, maybe, choosing c=0 for both cases was perhaps premature

Office_Shredder said:
When you plot both of them, what exactly is the difference? Try expanding the second function, and see if, maybe, choosing c=0 for both cases was perhaps premature

As it happens, by adding 2.25 to the latter, the both graphs coincide with each other. How come? Why is the first method easier to get to one primitive function directly?

Are you using the same C in each case? (Set x = 0 and see.)

there's no mistake. when you integrate a function, u get a constant C. this C can be any value, as any constant when differentiated gives you 0. so u can't merely take C = 0 and plot the graph that way. that's why they coincide when you add 2.25.

Mattara said:
As it happens, by adding 2.25 to the latter, the both graphs coincide with each other. How come? Why is the first method easier to get to one primitive function directly?

The first method is easier because integrating polynomials is easy. The second method is much more useful when you have stuff like x*cos(x2), and you can't just expand it and integrate.

I see.

By expanding

$$\frac {(4x-3)^3}{12}$$

I get

$$5 \frac {1}{3}x^3 -4x^2 + 3x + 2.25$$

And the constant is indeed 2.25.

So in other words, both are equally correct with respect to the initial problem (find one primitive function to $$(4x-3)^2$$)?

Both are a primitive function, only with different C values (0 and 2.25 resp.)?

No... you can't just throw the constant value away. Your two answers SHOULD look like

$$\frac {(4x-3)^3}{12} + C$$

and

$$\frac {16x^3}{3} - 12x^2 + 9x + C$$

What normally happens is you're given more information, e.g. what f(0) is, to determine what C actually is (it's not the same for the two different equations). But you can't just toss the constant out, since you're ignoring the complete set of solutions

I see.

Does that means that a satisfactory answer to the problem

"Find one primitive function to $$(4x-3)^2$$"

is

$$\frac {16x^3}{3} - 12x^2 + 9x$$

or

$$\frac {(4x-3)^3}{12}$$

or any of the two with a specified C value (ie. 1,2,3,4...)?

where as if it had been "Find all primitive functions to $$(4x-3)^2$$", the answer would have been

$$\frac {(4x-3)^3}{12} + C$$

and

$$\frac {16x^3}{3} - 12x^2 + 9x + C$$

?

Last edited:
Mattara said:
I see.

Does that means that a satisfactory answer to the problem

"Find one primitive function to $$(4x-3)^2$$"

is

$$\frac {16x^3}{3} - 12x^2 + 9x$$

or

$$\frac {(4x-3)^3}{12}$$

or any of the two with a specified C value (ie. 1,2,3,4...)?

where as if it had been "Find all primitive functions to $$(4x-3)^2$$", the answer would have been

$$\frac {(4x-3)^3}{12} + C$$

and

$$\frac {16x^3}{3} - 12x^2 + 9x + C$$

?
Yes, that's correct. $$\frac{(4x-3)^3}{12}$$ is one primitive function, where C=0. Likewise for your polynomial. $$\frac{(4x-3)^3}{12}+C$$ defines the entire family of primitive functions, where C can take any value.

## What is a primitive function?

A primitive function, also known as an antiderivative, is the inverse operation of a derivative. It is a function that, when differentiated, gives the original function.

## Why is it important to find a primitive function?

Finding a primitive function allows us to solve integrals, which are used in many real-world applications such as calculating areas, volumes, and work done.

## What is the process for integrating (4x-3)^2?

To integrate (4x-3)^2, we use the power rule for integration, which states that the integral of x^n is (x^(n+1))/(n+1) + C. In this case, n=2, so the integral becomes (4x-3)^3/3 + C.

## Can we use any other methods to integrate (4x-3)^2?

Yes, we can also use substitution or integration by parts to integrate (4x-3)^2. However, the power rule is the most straightforward method for this particular function.

## What is the general formula for integrating a polynomial function?

The general formula for integrating a polynomial function is to use the power rule for integration, where the integral of x^n is (x^(n+1))/(n+1) + C. We can apply this rule to each term in the polynomial and add the resulting integrals together to get the overall primitive function.

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