Why Do I Get Negative Value Integrating y=-50e^-5x?

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Discussion Overview

The discussion revolves around the definite integral of the function $$y=-50e^{-5x}$$ and the confusion regarding the negative value obtained from the integral despite the graphical representation suggesting otherwise. Participants explore the implications of integrating a function that is always negative over a specified interval.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the function $$-50e^{-5x}$$ is always negative for $$x$$ in the range from 0 to 1, leading to a negative definite integral.
  • Another participant questions the interpretation of the area under the curve, suggesting confusion about how the integral relates to the area bounded by the function and the axes.
  • A third participant points out that the integral function itself is positive but decreasing, which could contribute to misunderstandings about the values obtained from the definite integral.
  • There is mention of the need to account for a constant of integration when graphing the indefinite integral, which may shift the graph downwards.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the integral's value and its graphical representation. There is no consensus on how to reconcile the negative integral with the graphical depiction of the area under the curve.

Contextual Notes

Some participants highlight the importance of understanding the relationship between the definite integral and the area under the curve, noting that the integral's value reflects the area below the x-axis, which may lead to confusion.

bergausstein
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Hello! And Good day!

I just want to ask why do I keep getting a negative value whenever I take a definite integral of function $$y=-50e^{-5x}$$ the graph is shown as the first image.

If you look at the graph of the integral of "y" there's no traceable negative value on the graph. Why is that? Please help me understand why. Thanks!
 

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bergausstein said:
I just want to ask why do I keep getting a negative value whenever I take a definite integral of function $$y=-50e^{-5x}$$ the graph is shown as the first image.

If you look at the graph of the integral of "y" there's no traceable negative value on the graph. Why is that? Please help me understand why. Thanks!
The first plot correctly shows the graph of the function $-50e^{-5x}$, which is always negative as $x$ goes from $0$ to $1$.

The second plot does not make sense at all, if you are trying to find the area of the region between the graph and the axis. That is given by a definite integral, so it is just a number, not a graph. In fact, $$\int_0^1-50e^{-5x}dx = \bigl[10e^{-5x}\bigr]_0^1 = 10(e^{-5} - 1) \approx -9.9326.$$ That answer is negative, corresponding to the fact that the region lies below the $x$-axis.
 
Opalg said:
The first plot correctly shows the graph of the function $-50e^{-5x}$, which is always negative as $x$ goes from $0$ to $1$.

The second plot does not make sense at all, if you are trying to find the area of the region between the graph and the axis. That is given by a definite integral, so it is just a number, not a graph. In fact, $$\int_0^1-50e^{-5x}dx = \bigl[10e^{-5x}\bigr]_0^1 = 10(e^{-5} - 1) \approx -9.9326.$$ That answer is negative, corresponding to the fact that the region lies below the $x$-axis.

Is it not that the area bounded by the function y and the x and y-axis corresponds to the y values of the integral of y? I''m still confused.
 
http://mathhelpboards.com/attachments/calculus-10/6521d1491653833-area-calculation-exponential2-png
First, note $\displaystyle f(x) = \int -50e^{-5x} \, dx = 10e^{-5x} + C$ Your plot above assumes $C = 0$. Fact is, it should be shifted down 10 units since $C=-10$ for the following reason ...

Consider the definite integral as a function ... $\displaystyle f(x) = \int_0^x -50e^{-5t} \, dt \implies f(0) = 0$ The y-values of the definite integral "accumulation" function are indeed negative. Graph is attached. Also note the evaluation of $f(1)$ calculated previously by Opalg using the FTC ...

View attachment 6544
 
The integral function is positive but decreasing. If b> a then f(b)< f(a) so that f(b)- f(a) is negative.
 

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