What is the net area between #f(x) = e^(3-2x)-2x+1# and the x-axis over #x in [0, 3 ]#?

Question

Evelyn Agard · Accepted Answer

The area of f(x) bounded by #x=0andx=3# equal

#A=A_1+A_2=13.798(unite)^2#

show the steps
The sketch of our function is

To find the point at which the curve intersects the axis of theX-axis

#f(x)=0#

#e^(3-2x)-2x+1=0#

#x=1.279#

#(1.279,0)#

the area equal

#A=A_1+A_2#

#A_1=int_a^bf(x)-(0)*dx#

#A_1=int_0^1.279e^(3-2x)-2x+1=[-e^(3-2*x)/2-x^2+x]_0^1.279=(9175435*e^3-20823629)/18350870=8.908019591376782(unite)^2#

#A_2=int_b^c(0)-f(x)#

#A_2=int_1.279^3(0)-(e^(3-2x)-2x+1)=[e^(3-2*x)/2+x^2-x]_1.279^3=(e^(-3)*(89281591*e^3+9175435))/18350870=4.89014466396688(unite)^2#

The area of f(x) bounded by #x=0andx=3# equal

#A=A_1+A_2=13.798(unite)^2#

Christopher Allers · Answer

undefined To find the net area between $ f(x) = e^{3-2x} - 2x + 1 $ and the x-axis over $ x $ in $[0, 3]$, you need to integrate the absolute value of the function within the given interval. Then, you can evaluate the integral. The net area can be expressed as:

$$ \text{Net Area} = \int_{0}^{3} |f(x)| \, dx $$

Substitute $ f(x) = e^{3-2x} - 2x + 1 $ into the integral, take the absolute value, and then integrate over the interval $[0, 3]$. After calculating the integral, you will get the net area between the function and the x-axis over the specified interval.