Is the integral of 2x + x + 1 equal to (1)(3) + \frac{1}{2}(1)(3) + 1?

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Homework Help Overview

The discussion revolves around evaluating the definite integral of the expression 2x + x + 1 from 1 to 2. Participants are examining the validity of a proposed equation involving the integral and specific numerical values.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the formulation of the integral and the numbers used in the proposed equation. There is a suggestion to clarify how the original poster arrived at those specific values. Some participants are exploring the simplification of the integral and considering alternative expressions.

Discussion Status

The discussion is active, with participants providing insights into the integral's evaluation and questioning the assumptions behind the original expression. There is acknowledgment of correct evaluations, but no explicit consensus on the original equation's formulation.

Contextual Notes

Some participants express uncertainty about the original equation's terms, particularly regarding whether the first term should be 2x or 2x^2. This indicates a potential misunderstanding or ambiguity in the problem setup.

courtrigrad
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Is [tex]\int^2_1 2x + x + 1 dx= (1)(3) + \frac{1}{2}(1)(3) + 1[/tex]?
 
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Why doncha tell us how u came up with those numbers...??

Daniel.
 
[tex]\int^2_1 2x + x + 1 dx = \int^2_1 3x + 1 dx = \left[\frac{3}{2}x^2 + x \right]^2_1[/tex]
When you solve that i think u get 5.5

Though i suspect that original equation would be
[tex]\int^2_1 2x^2 + x + 1 dx[/tex] or why would they give it to you in that form ... 2x + x ?
 
Last edited:
They were trying to illustrate the basic integrals

[tex]\int^b_a x dx = \frac{1}{2}(b-a)(b+a)[/tex] and [tex]\int^b_a x^2 dx = \frac {1}{3}(b^3 - a^3)[/tex]

[tex]\int^b_a c dx = c(b-a)[/tex]

This is how I got those numbers
 
You answer is correct.
 
your answer is absolutely correct. (if you sure the first term is 2x instead of 2x^2 as ryoukomaru stated earlier)
 

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