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- Mathematica
- Thread starter Mr Davis 97
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- #1

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- #2

tnich

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Looks like you have only specified the lower limits of the sum and product (1), but no the upper limits (3).

- #3

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Oh. I meant int (Product[x+r, {r, 3}])*(1+ x*(Sum[1/(x+r), {r, 3}])) dx from 0 to 1. But this doesn't work either... Everything works when I remove that x right by the sum, but doesn't work when I add it in.Looks like you have only specified the lower limits of the sum and product (1), but no the upper limits (3).

- #4

tnich

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Try putting the ##x## inside the sum.Oh. I meant int (Product[x+r, {r, 3}])*(1+ x*(Sum[1/(x+r), {r, 3}])) dx from 0 to 1. But this doesn't work either... Everything works when I remove that x right by the sum, but doesn't work when I add it in.

- #5

tnich

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It looks to me like the integral doesn't converge. Try doing the integral analytically (using the calculus) instead of numerically (using Mathematica) and see what you get.Try putting the ##x## inside the sum.

- #6

tnich

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Now I'm not sure that it doesn't converge, but still, multiply out the terms in the integral, do the integration, and see what you get. Then you will know if your are getting the right answer from Mathematica.It looks to me like the integral doesn't converge. Try doing the integral analytically (using the calculus) instead of numerically (using Mathematica) and see what you get.

- #7

lurflurf

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Try one of these

Integrate[x (x + 1) (x + 2) (x + 3) (1/x + 1/(x + 1) + 1/(x + 2) + 1/(x + 3)), {x, 0, 1}]

Integrate[Product[x + y, {y, 3}] (1 + x Sum[1/(x + z), {z, 3}]), {x, 0, 1}]

Integrate[Sum[Product[x + y, {y, 0, 3}]/(x + z), {z, 0, 3}], {x, 0, 1}]

Integrate[Product[x + y, {y, 3}] Sum[x/(x + z), {z, 0, 3}], {x, 0, 1}]

Integrate[Sum[Pochhammer[x, 4]/(x + z), {z, 0, 3}], {x, 0, 1}]

Does anyone know a shorter way than five?

int (Product[x + r, {r, 1}])*(1 + x*(Sum[1/(x + r), {r, 1}])) dx

Integrate[Product[x + r, {r, 3}] (1 + x*(Sum[1/(x + r), {r, 3}])), {x, 0, 1}]

is the closest Mathematica form to what you wrote Wolframalpha has less strict syntax.

- #8

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$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,

Fred

- #9

tnich

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I get the same integrand, but 24 for the numerical result.

$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,

Fred

- #10

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Your right--my bad. I'm getting too old to integrate polynomials in my head.I get the same integrand, but 24 for the numerical result.

Peace,

Fred

- #11

lurflurf

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You used a numerical approach as 20 is a (wrong) number.

$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,

Fred

Integration by parts or substitution can be used to avoid expanding the polynomial.

That reminds me of a nice thing

general result

Differences[Pochhammer[{0, 1}, n]]

Integrate[D[Pochhammer[x, n], x], {x, 0, 1}]

##\int_0^1\mathrm{d}(x)_n=(1)_n-(0)_n=n!##

specific n=4

Differences[Pochhammer[{0, 1}, 4]]

Integrate[D[Pochhammer[x, n], x], {x, 0, 1}]

##\int_0^1\mathrm{d}(x)_4=(1)_4-(0)_4=4!##

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