Integral Evaluation Using u-Substitution: x*sqrt(1+x)dx from 0 to 8

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Homework Help Overview

The discussion revolves around evaluating the integral of the function x*sqrt(1+x) using u-substitution, specifically from the limits 0 to 8. The subject area is integral calculus, focusing on techniques for integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up the u-substitution but expresses confusion regarding the manipulation of terms, particularly the presence of a negative one in their expression. Some participants suggest a simpler approach to the substitution, emphasizing the need to derive u directly without squaring it.

Discussion Status

The discussion is ongoing, with participants providing alternative perspectives on the u-substitution method. There is a focus on clarifying the approach rather than reaching a consensus on a specific solution.

Contextual Notes

Participants are navigating the complexities of the substitution process and the definitions of polynomial versus other forms of expressions, indicating a level of uncertainty in the original poster's setup.

Destroxia
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Homework Statement


Evaluate the integral using usub

int(x*sqrt(1+x)dx) from 0 to 8

Homework Equations

The Attempt at a Solution


Okay, so I attempted to set for u sub

U = x+1
u^2 = (x+1)^2
2u*du = 2(x+1)*dx
u*du -1 = x*dx

I got this far but I don't understand what to do with this, because of the (-1), separate from the u*du
 
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There is no reason to square u.
Just take the derivatives, du = dx, x = u-1. Then you have a normal polynomial in u.
 
RUber said:
There is no reason to square u.
Just take the derivatives, du = dx, x = u-1. Then you have a normal polynomial in u.

Slight nit-pick -- it isn't actually a polynomial though it does only have terms of the form ##x^p##.
 
LCKurtz said:
Slight nit-pick -- it isn't actually a polynomial though it does only have terms of the form ##x^p##.

In the "Geometric Programming" literature, it would be called a "signomial". (It would further be called a "posynomial" if all its coefficients were ##>0##.)
 

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