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    Physics Pressure Data - Area under a Curve by Integration

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    (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.

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    Solution Preview

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    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.)