# Integral calculus problem

1. Jun 7, 2012

### wonglk9090

Deleted

None

Last edited: Jun 7, 2012
2. Jun 7, 2012

### algebrat

What class is this for, where did the problem come from. A text or written by prof?

3. Jun 7, 2012

### wonglk9090

That is just Cal 1 class in college level...I guess it is written by my professor.
Is that too difficult??lol

4. Jun 7, 2012

### Millennial

This is plain impossible. The solution can be expressed in terms of the error function if it was negative x squared (which is not elementary):

$$\displaystyle \frac{\mathrm{erf}(3)\sqrt{\pi}}{2}$$

But since it is positive, we need to use the Dawson integral function. The answer then becomes

$$\displaystyle e^9\mathrm{Di}(3)$$

where Di denotes the Dawson integral.

Last edited: Jun 7, 2012
5. Jun 7, 2012

### HallsofIvy

I suggest you reread the problem and then, possibly, ask your teacher if there has not been a typo. If the function were $xe^{x^2}$, this would be easy- and . But the function $e^{x^2}$ cannot be integrated in terms of elementary functions.

6. Jun 7, 2012

### meldraft

Your professor is sneaky, you really must make use of all the data:

$$\int f(x) dx=x f(x)-\int f'(x)xdx=xf(x)-\int e^{x^2}(\frac{x^2}{2})'dx=xf(x)-\frac{1}{2} \int (e^{x^2})' dx=xf(x)-\frac{e^{x^2}}{2}+c$$

Now, since this is a definite integral:

$$\int_0^3 f(x) dx=3f(3)-\frac{e^{3^2}}{2}-0f(0)+\frac{e^{0^2}}{2}=21-\frac{e^{3^2}}{2}+\frac{1}{2}$$