Thomas E.

asked • 01/28/18# This was on my first Calc 2 quiz: $2\int_{-3}^{3}e^x^4dx$

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.

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## 2 Answers By Expert Tutors

David D. answered • 07/12/18

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Former Community College Professor misses Teaching and Tutoring

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

I=∫

_{-3}^{3}e^{x^4}dx=2∫_{0}^{3}e^{x^4}dx=|x2=t, dx=dt/(2√t), 0≤t≤9|=∫_{0}^{9}(e^{t^2}/√t)dt.I

^{2}=∫_{0}^{9}∫_{0}^{9}e^{(x^2+y^2)}/√(xy)dxdy.Now make substitution x=ρcos(φ), x=ρsin(φ). Then

I

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

_{0}^{6/√π}e^{ρ^2}∫_{0}^{2π}(1/√(sin(2φ))dφ.Now, the first integral can be expressed through Error function and the second as Elliptic function.

David D.

I believe this solution is incorrect!

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07/12/18

Bobosharif S.

tutor

I might agree with you as I see the correct solution.

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07/12/18

David D.

The function e^x^4 has no closed form anti-derivative. That is equivalent to what you said

RE: an error function or an elliptic integral. I didn't see your remarks at the end. My apologies.

They could/should have been made at the end of the first line when the e^t^2 showd up. All the effort with spherical coordinates was merely stretching muscles. The result was obvious even before the first step. So, perhaps algebraically correct I thought for a moment that you had found an explicit anti-derivative(?) Since it was the first quiz in Calc II, I think the problem is obviously missing an x^3 term which will make it a u-substitution, an easy one. One could integrate by parts a couple times but you never get a "good vdu term to integrate. So why bother with all the fluff?

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07/12/18

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Bobosharif S.

^{x^4}?01/28/18