What is the net area between #f(x) = 3x^2-4x+2# and the x-axis over #x in [1, 3 ]#?

Question

Emilia Aguilar · Accepted Answer

#int_1^3 (3x^2-4x+2) dx=14#

To work this out, we need to evaluate the definite integral between 3 and 1.

Area=#int_1^3 (3x^2-4x+2) dx# #=[3(x^3/3)-4(x^2/2)+2x]_1^3# #=[x^3-2x^2+2x]_1^3# #=[(3)^3-2(3)^2+2(3)]-[(1)^3-2(1)^2+2(1)]# #=15-1# #=14#

Paisley Johnson · Answer

undefined To find the net area between $ f(x) = 3x^2 - 4x + 2 $ and the x-axis over $ x $ in the interval $[1, 3]$, you need to integrate the absolute value of the function from $ x = 1 $ to $ x = 3 $.

The function intersects the x-axis at points where $ f(x) = 0 $. To find these points, solve the equation $ 3x^2 - 4x + 2 = 0 $ for $ x $. The solutions are $ x = 1 $ and $ x = 2/3 $. However, since the interval is $[1, 3]$, we only need to consider $ x = 1 $ as the lower bound.

Integrating the absolute value of $ f(x) $ from $ x = 1 $ to $ x = 3 $ gives:

$$ \int_{1}^{3} |3x^2 - 4x + 2| dx $$

Split the interval $[1, 3]$ into two intervals: $[1, 2/3]$ and $[2/3, 3]$, and integrate separately.

For the interval $[1, 2/3]$, $ |3x^2 - 4x + 2| = 3x^2 - 4x + 2 $.

For the interval $[2/3, 3]$, $ |3x^2 - 4x + 2| = -(3x^2 - 4x + 2) $.

Calculate both integrals:

$$ \int_{1}^{2/3} (3x^2 - 4x + 2) dx $$
$$ \int_{2/3}^{3} -(3x^2 - 4x + 2) dx $$

Then, sum the absolute values of both results to find the net area between the function and the x-axis over the given interval.