What is the solution to the integral without an equation?

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Discussion Overview

The discussion revolves around the evaluation of a definite integral involving a function \( p(x) \) over the interval from 0 to 4. Participants are attempting to understand the calculations and reasoning behind the areas represented by the integral, particularly in segments of the interval.

Discussion Character

  • Mathematical reasoning, Homework-related, Technical explanation

Main Points Raised

  • One participant questions how the areas from 0 to 2 and from 2 to 4 are determined, suggesting that one cannot simply assume the area is half.
  • Another participant provides a calculation showing that the integral from 0 to 4 can be split into two parts, leading to a specific numerical result.
  • A further response reiterates the property of integrals that allows for the addition of integrals over adjacent intervals, although it does not clarify the specific areas in question.
  • One participant presents a detailed breakdown of the integral, including terms that involve \( p(x) \) and constants, leading to a different numerical outcome.

Areas of Agreement / Disagreement

The discussion reflects multiple competing views regarding the evaluation of the integral and the assumptions about the areas involved. There is no consensus on the correct approach or final answer.

Contextual Notes

Participants express uncertainty about the assumptions made regarding the areas under the curve and the specific properties of the function \( p(x) \). There are unresolved mathematical steps in the calculations presented.

Who May Find This Useful

Students or individuals interested in integral calculus, particularly those grappling with definite integrals and area calculations.

karush
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I didn't see how they got ii and iii
Bk answers are in red
 
Last edited:
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$$\int_0^4 (2-p(x)) dx= \int_0^4 2 dx - \int_0^4 p(x) dx= 2 \cdot 4-6=2$$
 
How do determine what the area 0-2 and 2-4 can't assume area is half?
For iii
 
Last edited:
karush said:
How do determine what the area 0-2 and 2-4 can't assume area is half?
For iii

What do you mean? I haven't understood your question.In general it holds that $\int_a^c f(x) dx+ \int_c^b f(x) dx= \int_a^b f(x) dx$.
 
$\int_{0}^{2}p\left(x\right) \,dx
+\int_{2}^{4}p\left(x\right) \,dx
-\int_{0}^{2}2 \,dx
+\int_{0}^{2}1 \,dx$

$\int_{0}^{4}p(x) \,dx-4+2$

$6-4+2=4$
 
Last edited:

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