This was on my first Calc 2 quiz: $2\int_{-3}^{3}e^x^4dx$

Question

I feel that there should be another x in the problem so that u sub works, or something, as it is I believe it is a partial gamma function or something, but I know we should not learn those until much later.

Answer 1

Thomas, you are probably correct about a missing term but I would guess the missing part is x^3. This makes the problem a simple u-sub.

 

u = x^4

1/3du=x^3

so int(x^e^x^4) = 1/3int(e^udu) = 1/2e^u =1/3e^x^4 evaluated from -3 to 3.

 

Got it?

 

Dr. Dave

Answer 2

I=∫_-3³e^{x^4}dx=2∫₀³e^{x^4}dx=|x2=t, dx=dt/(2√t), 0≤t≤9|=∫₀⁹(e^{t^2}/√t)dt.

I²=∫₀⁹∫₀⁹e^{(x^2+y^2)}/√(xy)dxdy.

Now make substitution x=ρcos(φ), x=ρsin(φ). Then

I²=√2∫₀^6/√π∫₀^2πe^{ρ^2}/√(sin(2φ)dρdφ

                           =√2∫₀^6/√πe^{ρ^2}∫₀^2π(1/√(sin(2φ))dφ. 

Now, the first integral can be expressed through Error function and the second as Elliptic function.