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

1 Answer
May 2, 2018

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

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

Explanation:

show the steps
The sketch of our function is

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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