Integral Fail

08142020, 03:44 AM
Post: #1




Integral Fail
The Prime and CASIO Classpad 2 both fail to integrate the indefinite integral
(3 NROOT x^2 2*x + 1)^4 The NSpire is successful, produces (3*(x1)^(11/3))/11 Any guidance on how to enter the integral differently so that the Prime can handle it? 

08142020, 01:06 PM
Post: #2




RE: Integral Fail
It's a bit cumbersome, but you have to do this:
factor(x^22*x1) > (x1)^2 Therefore: factor(int(((x1)^2)^(4/3),x)) > (3*(x1)^3*((x1)^(1/3))^2)/11 Welcome those who have a less cumbersome answer! 

08142020, 01:18 PM
Post: #3




RE: Integral Fail
I apologize:
factor(x^22*x+1) > (x1)^2 Therefore: factor(int(((x1)^2)^(4/3),x)) > (3*(x1)^3*((x1)^(1/3))^2)/11 

08142020, 01:34 PM
Post: #4




RE: Integral Fail
The problem is that the Prime doesn't do the factorization itself.
Arno 

08142020, 01:50 PM
Post: #5




RE: Integral Fail
Perhaps the peoblem comes from xCas...


08142020, 03:47 PM
(This post was last modified: 08142020 04:08 PM by Albert Chan.)
Post: #6




RE: Integral Fail
(08142020 03:44 AM)lrdheat Wrote: The Prime and CASIO Classpad 2 both fail to integrate the indefinite integral However, the "successsful" result were wrong. f(x) = x1 ^ (8/3) If x ≥ 1, F(x)  F(1) = 3/11 * x1^(11/3) If x < 1, F(x)  F(1) = −3/11 * x1^(11/3) → F(x) = 3/11 * (x1) * x1^(8/3) + C XCas> F := int(abs(x1)^(8/3),x) XCas> F := factor(F) → \(\large \frac{3 \left(x1\right)^{3} \left(\left((x1) \mathrm{sign}\left(x1\right)\right)^{\frac{1}{3}}\right)^{2}}{11}\) 

08152020, 01:44 AM
Post: #7




RE: Integral Fail
Hi Albert,
Are you sure about this? When I do this problem using numerical integration On the Prime, Classpad, and TI30X Pro Math Print, I get 3.46+ which equals the “correct” NSpire result. This does not agree with your result (unless I am doing something wrong...). 

08152020, 02:03 AM
Post: #8




RE: Integral Fail
My HP 50g produces 3.46...


08152020, 04:38 AM
Post: #9




RE: Integral Fail
Here comes the problem that always puts me in difficulty. If I set CAS without "use i to factor polynomials" I get:
sqrt((x1)^2) > abs(x1) and the integral int(abs(x1)^(8/3),x) takes the form proposed by Albert Chan. However, if I set CAS with "use i to factor polynomials", the result takes the form of "robmio" and, for negative "x" values, the result of the integral gives a complex number. So what is the correct result? 

08152020, 04:45 AM
Post: #10




RE: Integral Fail
(08152020 01:44 AM)lrdheat Wrote: Hi Albert, To reproduce your result, I am assuming integral limit from 1 to 1 XCas> gaussquad((x^22*x+1)^(4/3), x=1 .. 1) → 3.46342047702 F(x) = 3/11 * (x1) * abs(x1)^(8/3) F(1)  F(1) =  F(1) = 3.46342047702 ... 

08152020, 05:20 AM
Post: #11




RE: Integral Fail
I’m sorry...I was integrating from 1 to 3. Do you get the same answer (that would imply symmetry about x=1). In the morning, I will check out the graph!


08152020, 11:44 AM
Post: #12




RE: Integral Fail
(08152020 05:20 AM)lrdheat Wrote: I was integrating from 1 to 3. Do you get the same answer (that would imply symmetry about x=1). Yes, f(x)=x1^(8/3) have symmetry at x=1. But, for F(x), we need to flip the sign. Because F(1)=0, integrating from 1 to 3 is same as 1 to 1 F(3)  F(1) = F(1+2) =  F(12) = F(1)  F(1) FYI, here is an equivalent F(x), using surd: XCas> F := int(surd(x1,3)^8,x) "Temporary replacing surd/NTHROOT by fractional powers" → \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x1)\right)^{2} \left(x1\right)^{3}\) // = \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x1)\right)^{11}\) 

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