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

1 Answer
May 23, 2017

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

Explanation:

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