Explore BrainMass

Physics Pressure Data - Area under a Curve by Integration

(See the 2 graphs in the attached question file)

I need to solve for the area under the 2 noted curves.

In the first graph, the beginning x,y coordinate is 0 minutes; 80 degrees F; the ending x,y coordinates are 4.5 minutes; 212 degrees F

In the second graph, the beginning x,y coordinate is 5.0 minutes; 213 degrees; the ending x,y coordinate is 20.0 minutes; 393 degrees F.

With this information and the 2 line equations given, please solve for the area under the curve and show me the steps in simple terms to solve.


Solution Preview

[Please scroll to the end for an Update.] ****

Graph 1:

Area under the curve = Integral (59.137 ln x + 82.276) dx over [0, 4.5]

= [Integral 59.137 ln x dx + Integral 82.276 dx] over [0, 4.5]

= [I1 + I2] over [0, 4.5]

Consider I1 = Integral 59.137 ln x dx = 59.137 * Integral ln x dx

We find the integral of ln x dx by integrating by parts, taking u as ln x and dv as dx

By ...

Solution Summary

The solution file contains the calculation of area under the curves by integration. (This is part of a Physics experiment on pressure.)