How can a series be equal to this integral?

[opus]

Homework Statement

Determine the series that is equal to the integral $\int_0^1 x^2\cos(x^3)dx$

Homework Equations

The Attempt at a Solution

So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be $\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}$

Now I have $\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx$
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have $\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx$
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after $\sum$ in brackets?

Eventually, I end up evaluating and I get $\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}$

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

 

[opus]

Also, apologies for the hideous notation. I don't really know what goes where, and this type of problem seems very "choppy" for me right now. So everything is pretty far from nice to look at

 

[opus]

Also again...
Here's an attachment of my work in case there were specific things that need to be addressed.

 

[Delta2]

...
Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why can't we have that two real numbers are equal?

 

[opus]

I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why can't we have that two real numbers are equal?

Here’s what I mean that goes along with the lat part of my first post that is part of my confusion. Not the “bargraph” like piece represents a series.

 

[Delta2]

Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that $f(x)=a_n , x\in[n,n+1)$ and the integral of that function from 0 to +infinite $\int_0^{+\infty}f(x)dx =\sum a_n$would be equal to the series .

 

[Mark44]

Don't overthink this.

Determine the series that is equal to the integral $\int_0^1 x^2\cos(x^3)dx$

Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!

 

[vela]

Eventually, I end up evaluating and I get
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}$$

Use the fact that $6n+3 = 3(2n+1)$. You should be able to identify the resulting series.

 

[opus]

Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that $f(x)=a_n , x\in[n,n+1)$ and the integral of that function from 0 to +infinite $\int_0^{+\infty}f(x)dx =\sum a_n$would be equal to the series .

Ok yes that answered my question, thank you.

 

[opus]

Don't overthink this.

Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!

So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.

Use the fact that $6n+3 = 3(2n+1)$. You should be able to identify the resulting series.

Thanks!

 

[Orodruin]

The series you have is not the Riemann sum of the integral. It is just a number of different contributions to it, integration is linear.

Compare with how you would integrate 1+x ...

 

[Delta2]

From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.

 

[Mark44]

So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.

I don't think the center is relevant here. For example, the Maclaurin series for $e^x$ is $1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots$, so it's straightforward to find, say, $e^1$.

From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity

Not relevant here. The integral in question is over the interval [0, 1].

, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.

 

[Ray Vickson]

Homework Statement

Determine the series that is equal to the integral $\int_0^1 x^2\cos(x^3)dx$

Homework Equations

The Attempt at a Solution

So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be $\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}$

Now I have $\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx$
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have $\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx$
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after $\sum$ in brackets?

Eventually, I end up evaluating and I get $\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}$

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

If your problem was to evaluate the indefinite integral $\int x^2 \cos(x^3) \, dx$ then, indeed, you would get an infinite series involving $x$. However, you have a definite integral in which $x = 1$, so that is why you don't see an $x$ in your answer. No rocket science involved here!

 

[Mark44]

If your problem was to evaluate the indefinite integral $\int x^2 \cos(x^3) \, dx$ then, indeed, you would get an infinite series involving $x$.

This is true, but somewhat misleading. Finding an antiderivative is very straightforward, and results in a simple function. As an additional step, you can rewrite that function as an infinite series.

However, you have a definite integral in which $x = 1$, so that is why you don't see an $x$ in your answer. No rocket science involved here!