Integral Fail - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Integral Fail (/thread-15466.html) |
Integral Fail - lrdheat - 08-14-2020 03:44 AM 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? RE: Integral Fail - robmio - 08-14-2020 01:06 PM 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! RE: Integral Fail - robmio - 08-14-2020 01:18 PM 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 RE: Integral Fail - Arno K - 08-14-2020 01:34 PM The problem is that the Prime doesn't do the factorization itself. Arno RE: Integral Fail - robmio - 08-14-2020 01:50 PM Perhaps the peoblem comes from xCas... RE: Integral Fail - Albert Chan - 08-14-2020 03:47 PM (08-14-2020 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) = |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) → \(\large \frac{3 \left(x-1\right)^{3} \left(\left((x-1) \mathrm{sign}\left(x-1\right)\right)^{\frac{1}{3}}\right)^{2}}{11}\) RE: Integral Fail - lrdheat - 08-15-2020 01:44 AM 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...). RE: Integral Fail - lrdheat - 08-15-2020 02:03 AM My HP 50g produces -3.46... RE: Integral Fail - robmio - 08-15-2020 04:38 AM 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? RE: Integral Fail - Albert Chan - 08-15-2020 04:45 AM (08-15-2020 01:44 AM)lrdheat Wrote: Hi Albert, 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 ... RE: Integral Fail - lrdheat - 08-15-2020 05:20 AM 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! RE: Integral Fail - Albert Chan - 08-15-2020 11:44 AM (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" → \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{2} \left(x-1\right)^{3}\) // = \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{11}\) |