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i don't even understand how to do this, surely this function is continuous at $|x|<1$ and have real value on that given integral. and $$\int_{-1}^{1}\sqrt{1-x^4}dx$$ should doable .

but since there isn't any variable outside root, i can't make any substitution of it. i'm trying sought on integral calculator but it said there isn't any antiderivative of this function .

so assuming this question is valid is there any way to solve this? ( i'm currently on mid-end of calculus 1). by the way the provided answer just attach half circular image that says "symmetry" and state $1.748$ is the answer and it doesn't make much sense for meenter image description here

pikarin-g
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2 Answers2

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In terms of Beta functions,$$\int_{-1}^1(1-x^4)^{1/2}dx=2\int_0^1(1-x^4)^{1/2}dx=\frac12\int_0^1y^{-3/4}(1-y)^{1/2}dy=\frac12\operatorname{B}\left(\frac14,\,\frac32\right).$$This can be rewritten with Gamma functions as$$\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac32\right)}{2\Gamma\left(\frac74\right)}=\frac{\Gamma\left(\frac14\right)\sqrt{\pi}}{3\Gamma\left(\frac34\right)}=\frac{\Gamma^2\left(\frac14\right)\sqrt{\pi}}{3\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)}=\frac{\Gamma^2\left(\frac14\right)}{3\sqrt{2\pi}}.$$

J.G.
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  • thank you i really appreciate that even i'm still far from there , and i'm attached the image of the worksheet . how your thought about it ? – pikarin-g Nov 18 '19 at 13:00
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It's an elliptic integral, which you are probably not expected to know anything about.

But note that $\sqrt{1-x^4}>\sqrt{1-x^2}$ for $-1<x<1$, and therefore $$ \int_{-1}^1 \sqrt{1-x^4} \, dx > \int_{-1}^1 \sqrt{1-x^2} \, dx = [\text{area of half the unit circle}] = \frac{\pi}{2} \approx 1.5708 . $$ So your integral is greater than $\pi/2$, and alternative (D) looks like a rounded value of $\pi/2$, and (E) is the only number which is larger, so the only alternative which can be correct is (E).

(Strictly speaking, it could be (D), since $1.571$ also also greater than $\pi/2$, but that possibility can be ruled out since the difference $1.571-\pi/2$ is so small that the total area between the graphs $y=\sqrt{1-x^4}$ and $y=\sqrt{1-x^2}$ must be greater.)

Hans Lundmark
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  • thanks a lot , since answer in worksheet is only make me confuse , it's keep highlighting on integral but not explain the nature of the function itself. – pikarin-g Nov 18 '19 at 13:48
  • @pikarin-g: Yes, if that handwritten stuff in the image you posted is supposed to be some kind of official answer key, it really doesn't provide much explanation...! – Hans Lundmark Nov 18 '19 at 13:58