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

Question

Ezra Adams · Accepted Answer

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#

Emily Ammerman · Answer

undefined To find the net area between $ f(x) = e^{3-x} - 2x + 1 $ and the x-axis over $ x $ in the interval $[1, 2]$, you need to integrate the absolute value of the function within that interval. The area can be computed as follows:

$$
\text{Net area} = \int_{1}^{2} |f(x)| \, dx
$$

Substitute $ f(x) = e^{3-x} - 2x + 1 $ and evaluate the integral over the given interval.

Paisley Johnson · Answer

undefined To find the net area between the function $ f(x) = e^{3-x} - 2x + 1 $ and the x-axis over the interval $ x $ in $[1, 2]$, you first need to compute the definite integral of the absolute value of the function over the given interval. Since the function $ f(x) $ can be negative within the interval, taking the absolute value ensures that the area is positive. Then, integrate the absolute value of the function over the interval $[1, 2]$ to find the net area.