Antiderivative of x^2sin(x^2) with Given Initial Condition

In summary, the problem states that find and antiderivative F of f(x)=x^2\sin(x^2) such that F(1)=0. The problem suggests using u=x^2 to simplify the problem, but this does not work and the product rule is used to detect the x^2. The solution is to add in a few functions that do not change f(x) and use the law of sine and cosine to find the antiderivative.
  • #1
Xcron
22
0
Problem states:

Find and antiderivative [tex]F[/tex] of [tex]f(x)=x^2\sin(x^2)[/tex] such that [tex]F(1)=0[/tex].

I thought at first that I would use [tex]u=x^2[/tex] to simplify the matter but I don't seem to get anywhere. I can't seem to figure out a way to start it off either...I was considering the Squeeze Theorem because they are functions, essentially.

Any advice/help?
 
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  • #2
Xcron said:
Problem states:
Find and antiderivative [tex]F[/tex] of [tex]f(x)=x^2\sin(x^2)[/tex] such that [tex]F(1)=0[/tex].
I thought at first that I would use [tex]u=x^2[/tex] to simplify the matter but I don't seem to get anywhere. I can't seem to figure out a way to start it off either...I was considering the Squeeze Theorem because they are functions, essentially.
Any advice/help?

How do you use the squeeze theorem to find an antiderivative? Try integration by parts. If you haven't learned integration by parts yet, then are you sure it isn't [tex] x\sin(x^2)[/tex]
 
  • #3
Nope...sorry, I haven't learned integration by parts yet...that will be covered in the second Calculus class that I will be taking in this Winter session at my college.

And yes, I am sure it isn't [tex]x\sin(x^2)[/tex]...that would be a completely superficial problem that I should not have any problems with =P.

I was thinking about this problem some more (been stuck on it since last night and it's almost the night of the next day haha) and I thought that I could possibly get an antiderivative that had [tex]\frac{1}{2}[/tex] in it but then I realized that no matter what I came up with...the product rule would ruin my plan. If only that [tex]x^2[/tex] was a constant haha...

About an hour ago, I thought about a method that I could use by adding in a few functions that wouldn't change the value of [tex]f(x)[/tex] and then I came up with a plan all of a sudden and it seemed to work in my head. I worked it over a few times in my head and realized that it may be the answer...I would have to use the law that states [tex]\sin^2(x)+\cos^2(x)=1[/tex].

Sadly, that didn't produce the answer either because I ended up with:
[tex]x^2\sin(x^2)=x^2\sin(x^2)(\sin^2(x^2)+\cos^2(x^2))=
x^2\sin^3(x^2)+x^2\sin(x^2)\cos^2(x^2)[/tex]

This didn't get me anywhere because the product rule caught me none the less...the [tex]x^2[/tex] was the problem in that case.

While typing all of this up, I decided to look in the back of the book to see what the answer is so that I could get some insight as to how to go about this problem...and I was a bit dumbfounded upon seeing it...I guess I forgot that an antiderivative can just be the general notation of an integral and the integrand...you don't necessarily need the actual antiderivative function...*sigh*...mmmm the answer was like this:

[tex]F(x)=\int_{1}^{x}t^2\sin(t^2)dt[/tex]
 
  • #4
That happened to me in a differential equations course once- I reduced the problem to an integral and spent hours trying to do the integral. The answer in the back of the book gave the solution in terms of the integral!:rolleyes:
 
  • #5
This is probably a properties of definite integrals question, cause there is no way you can integrate this that I know of without using IBP.
 
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  • #6
[tex] \int x^{2}\sin x^{2}dx= -\frac{1}{2}\left( \cos x^{2}\right) x+\frac{1}{4}\sqrt{2}\sqrt{\pi }\func{FresnelC}\left( \frac{\sqrt{2}}{\sqrt{\pi }}x\right) + C [/tex]

Well, set x=1 in the RHS and equal it to 0.

Daniel.
 

What is an antiderivative?

An antiderivative is a mathematical function that is the inverse operation of a derivative. It is the original function that, when differentiated, gives the given function.

How do you find an antiderivative?

To find an antiderivative, you can use the process of integration. This involves using various techniques, such as u-substitution or integration by parts, to find the original function.

Why is finding antiderivatives important?

Finding antiderivatives is important because it allows us to solve problems involving rates of change and to find the original function from a given derivative. This is useful in many real-world applications, such as physics, engineering, and economics.

What is the difference between an antiderivative and an indefinite integral?

An antiderivative is a single function that is the inverse operation of a derivative, while an indefinite integral is a family of functions that differ only by a constant. In other words, an antiderivative gives a specific solution, while an indefinite integral gives a set of solutions.

What are some common techniques for finding antiderivatives?

There are several techniques for finding antiderivatives, including u-substitution, integration by parts, partial fractions, and trigonometric substitutions. It is important to be familiar with these techniques and know when to use them in order to successfully find antiderivatives.

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