Integral Fail
08-14-2020, 03:44 AM
Post: #1
 lrdheat Senior Member Posts: 779 Joined: Feb 2014
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*(x-1)^(11/3))/11

Any guidance on how to enter the integral differently so that the Prime can handle it?
08-14-2020, 01:06 PM
Post: #2
 robmio Member Posts: 164 Joined: Jan 2020
RE: Integral Fail
It's a bit cumbersome, but you have to do this:

factor(x^2-2*x-1) -> (x-1)^2

Therefore:

factor(int(((x-1)^2)^(4/3),x)) -> (3*(x-1)^3*((x-1)^(1/3))^2)/11

Welcome those who have a less cumbersome answer!
08-14-2020, 01:18 PM
Post: #3
 robmio Member Posts: 164 Joined: Jan 2020
RE: Integral Fail
I apologize:

factor(x^2-2*x+1) -> (x-1)^2

Therefore:

factor(int(((x-1)^2)^(4/3),x)) -> (3*(x-1)^3*((x-1)^(1/3))^2)/11
08-14-2020, 01:34 PM
Post: #4
 Arno K Senior Member Posts: 463 Joined: Mar 2015
RE: Integral Fail
The problem is that the Prime doesn't do the factorization itself.
Arno
08-14-2020, 01:50 PM
Post: #5
 robmio Member Posts: 164 Joined: Jan 2020
RE: Integral Fail
Perhaps the peoblem comes from xCas...
08-14-2020, 03:47 PM (This post was last modified: 08-14-2020 04:08 PM by Albert Chan.)
Post: #6
 Albert Chan Senior Member Posts: 2,101 Joined: Jul 2018
RE: Integral Fail
(08-14-2020 03:44 AM)lrdheat Wrote:  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*(x-1)^(11/3))/11

Any guidance on how to enter the integral differently so that the Prime can handle it?

However, the "successsful" result were wrong.

f(x) = |x-1| ^ (8/3)

If x ≥ 1, F(x) - F(1) = ﻿ ﻿ 3/11 * |x-1|^(11/3)
If x < 1, F(x) - F(1) = 3/11 * |x-1|^(11/3)

→ F(x) = 3/11 * (x-1) * |x-1|^(8/3) + C

XCas> F := int(abs(x-1)^(8/3),x)
XCas> F := factor(F)

→ $$\large \frac{3 \left(x-1\right)^{3} \left(\left((x-1) \mathrm{sign}\left(x-1\right)\right)^{\frac{1}{3}}\right)^{2}}{11}$$
08-15-2020, 01:44 AM
Post: #7
 lrdheat Senior Member Posts: 779 Joined: Feb 2014
RE: Integral Fail
Hi Albert,

Are you sure about this? When I do this problem using numerical integration On the Prime, Classpad, and TI-30X 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...).
08-15-2020, 02:03 AM
Post: #8
 lrdheat Senior Member Posts: 779 Joined: Feb 2014
RE: Integral Fail
My HP 50g produces -3.46...
08-15-2020, 04:38 AM
Post: #9
 robmio Member Posts: 164 Joined: Jan 2020
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((x-1)^2) -> abs(x-1)

and the integral

int(abs(x-1)^(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?
08-15-2020, 04:45 AM
Post: #10
 Albert Chan Senior Member Posts: 2,101 Joined: Jul 2018
RE: Integral Fail
(08-15-2020 01:44 AM)lrdheat Wrote:  Hi Albert,

Are you sure about this? When I do this problem using numerical integration On the Prime, Classpad, and TI-30X 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...).

To reproduce your result, I am assuming integral limit from -1 to 1

XCas> gaussquad((x^2-2*x+1)^(4/3), x=-1 .. 1) ﻿ ﻿ ﻿ ﻿ → 3.46342047702

F(x) = 3/11 * (x-1) * abs(x-1)^(8/3)

F(1) - F(-1) = - F(-1) = 3.46342047702 ...
08-15-2020, 05:20 AM
Post: #11
 lrdheat Senior Member Posts: 779 Joined: Feb 2014
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!
08-15-2020, 11:44 AM
Post: #12
 Albert Chan Senior Member Posts: 2,101 Joined: Jul 2018
RE: Integral Fail
(08-15-2020 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)=|x-1|^(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(1-2) = F(1) - F(-1)

FYI, here is an equivalent F(x), using surd:

XCas> F := int(surd(x-1,3)^8,x)
"Temporary replacing surd/NTHROOT by fractional powers"

→ $$\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{2} \left(x-1\right)^{3}$$ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// = $$\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{11}$$
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