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

[bergausstein]

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!

 

[Opalg]

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.

 

[bergausstein]

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.

 

[skeeter]

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

 

[HallsofIvy]

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