Negative area between two curves

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SUMMARY

The discussion centers on calculating the area between the curves defined by the functions f(x) = 3^x and g(x) = 2x + 1. The intersections occur at x = 0 and x = 1, leading to the integral from 0 to 1 of (g(x) - f(x)) dx, which results in a negative area of approximately -0.1795. The negative value indicates that the function 3^x - (2x + 1) is predominantly negative within the interval, confirming that the upper function was incorrectly identified. Correctly integrating (2x + 1 - 3^x) yields a positive area, clarifying the importance of identifying the upper and lower functions accurately.

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adriaat
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I've been trying to figure out what a negative area means, but I can't.

Homework Statement
Calculate the area between f(x) = 3^{x} \, , \, g(x)=2x+1

The attempt to a solution
The intersections are located in x=0 and x=1.
So I do the integral from 0 to 1.
\int_{0}^{1} (g(x)-f(x))dx = \frac{2}{log(3)}-2 \approx -0.17952154675

What am I doing wrong? I am integrating the upper function minus the lower function.
 
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The Negative is only telling that the function 3^x-2x-1,has negative values more than positive ones inside the interval of integration and of course by integrating 2x+1-3^x you will get a positive answer!It is obvious from this that the sign doesn't mean much.
 
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adriaat said:
\int_{0}^{1} (g(x)-f(x))dx = \frac{2}{log(3)}-2
No, you have swapped f(x) and g(x) in this step.
 
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