What is the net area between #f(x) = e^(3-x)-2x+1# and the x-axis over #x in [1, 2 ]#?

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Answer 1 · 2017-05-23T22:36:55

The net area is #e^2-e-2 approx 2.6708#

Net area between the graphs is the absolute value of the integral between the limits of the integral:

#| int_1^2 (e^(3-x) -2x + 1) dx |#

Evaluate the definite
# = | [-e^(3-x) - x^2 + x]_1^2 |#

# = | (-e^1 - 4 + 2) - (-e^2-1+1) |#

# = | -e -2 + e^2|#

Because the value inside the absolute value is positive, this can be simplified to:
# = e^2-e-2#

# approx 2.6708#