Calc 3, taylor..: int[0,x+y^2] exp(−t^2)dt. degree 3 near (0,0).

[Knorrhane]

[FONT=wf_segoe-ui_normal]\(\displaystyle \int_{o}^{x+y^2} \exp(-t^2) dt\)[/FONT][FONT=wf_segoe-ui_normal]. degree 3 near (0,0). And I get it to \(\displaystyle x+y^2-\frac{(x+y^2)^3}{3}\) while answer is suppose to be \(\displaystyle x+y^2-\frac{x^3}{3}\). Is it something I'm missing?[/FONT]

 

[stapel]

\(\displaystyle \displaystyle \int_{0}^{x+y^2}\, e^{-t^2}\, dt,\, \mbox{ degree 3 near }\, (0,\, 0)\)

And I get it to:

. . .\(\displaystyle x\,+\,y^2\,-\,\dfrac{(x\,+\,y^2)^3}{3}\)

while answer is suppose to be:

. . .\(\displaystyle x\,+\,y^2\,-\,\dfrac{x^3}{3}\)

Is it something I'm missing?

Please reply showing your work. Thank you!

 

[Knorrhane]

Please reply showing your work. Thank you!

So McLaurin series for e^x is \(\displaystyle 1+x+\frac{x^2}{2}+....\) switch all x for -t^2.

\(\displaystyle \int_{0}^{x+y^2} (1-t^2)dt = t-\frac{t^3}{3}\) plug in x+y^2 for all t's, yields \(\displaystyle x+y^2-\frac{(x+y^2)^3}{3}\)

 

[HallsofIvy]

You have not followed the directions of the problem correctly. What you have is
\(\displaystyle x+ y^2+ \frac{(x+ y^2)^3}{3}= x+ y^2+ \frac{x^3}{3}+ \frac{3x^2y^2}{3}+ \frac{3xy^4}{3}+ \frac{y^6}{3}\)
and you were asked to find a polynomial of degree 3.

 

[Knorrhane]

You have not followed the directions of the problem correctly. What you have is
\(\displaystyle x+ y^2+ \frac{(x+ y^2)^3}{3}= x+ y^2+ \frac{x^3}{3}+ \frac{3x^2y^2}{3}+ \frac{3xy^4}{3}+ \frac{y^6}{3}\)
and you were asked to find a polynomial of degree 3.

Aha, I thought \(\displaystyle t^3\) was my 3rd degree. Thanks for clarifying