How do I set this antiderivative up?

  • #1
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Mod note: Moved from technical math section, so there's no template.
Sorry if I'm formatting this question wrong, new user.
F(x) is an antiderivative of [PLAIN]https://upload.wikimedia.org/math/9/1/5/915ca58b070b0328cd069524c2d487f2.pngx[SUP]3[/SUP]+x+1. [Broken] F(1)https://upload.wikimedia.org/math/4/3/e/43ec3e5dee6e706af7766fffea512721.png -2.125. Find F(4).
I tried to set up the antiderivative as (2/3)(x^3+x+1)^(3/2) and then plugging 1 in to get -2.125, but its clearly wrong. Is there another step I need to do?
 
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Answers and Replies

  • #2
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posted in the wrong place sorry thought this was the homework forum. how do i delete post?
 
  • #3
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Sorry if I'm formatting this question wrong, new user.
F(x) is an antiderivative of [PLAIN]https://upload.wikimedia.org/math/9/1/5/915ca58b070b0328cd069524c2d487f2.pngx[SUP]3[/SUP]+x+1. [Broken] F(1)https://upload.wikimedia.org/math/4/3/e/43ec3e5dee6e706af7766fffea512721.png -2.125. Find F(4).
I tried to set up the antiderivative as (2/3)(x^3+x+1)^(3/2)
No, that isn't right, assuming that the integrand is ##\sqrt{x^3 + x + 1}##. If you differentiate your proposed antiderivative, you don't get ##\sqrt{x^3 + x + 1}##.
jakka said:
and then plugging 1 in to get -2.125, but its clearly wrong. Is there another step I need to do?

Are you sure you have copied the problem correctly? As written this is not a simple integration problem.
 
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  • #4
Ray Vickson
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Mod note: Moved from technical math section, so there's no template.
Sorry if I'm formatting this question wrong, new user.
F(x) is an antiderivative of [PLAIN]https://upload.wikimedia.org/math/9/1/5/915ca58b070b0328cd069524c2d487f2.pngx[SUP]3[/SUP]+x+1. [Broken] F(1)https://upload.wikimedia.org/math/4/3/e/43ec3e5dee6e706af7766fffea512721.png -2.125. Find F(4).
I tried to set up the antiderivative as (2/3)(x^3+x+1)^(3/2) and then plugging 1 in to get -2.125, but its clearly wrong. Is there another step I need to do?

If you mean ##f(x) = \sqrt{x^3+x+1}## then ##F(a) = \int_0^a f(x) \, dx## evaluates to a horrible expression involving Elliptic functions. For the record, here is what I get when I do the integral using the computer algebra package Maple:
.400000000000000*a*(a^3+a+1.)^(1/2)+.747347077668721+.334696371293003e-1*(1944.39383604376-905.253833500497*a+1326.71397592448*a^2)^(1/2)/(6.33800237600924+9.28879394984575*a)^(1/2)-.760562326439714*EllipticF(29.7401409537626*(6.33800237600924+9.28879394984575*a)^(1/2)/(81.2424695605669+36.4240851075835*a),.911348956138490)+.995391725896672*EllipticE(29.7401409537626*(6.33800237600924+9.28879394984575*a)^(1/2)/(81.2424695605669+36.4240851075835*a),.911348956138490)+.283999288535776/(81.2424695605669+36.4240851075835*a)*(-.536375634160795e-1*(382.891652874454*a-221.118995987414*a^2-165.754662083937)*(2061.23266890537*a^2-1406.43636015277*a+3020.88330175229))^(1/2)/(6.33800237600924+9.28879394984575*a)^(1/2)
 
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