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

Question

Emily Ammerman · Accepted Answer

#1/3e^6-1/3e^3-6~=121.78# We can represent this area as the integral of #f(x)# on the interval #[1,2]#: #int_1^2e^(3x)-4x\ dx# We start by computing the antiderivative: #int\ e^(3x)-4x\ dx=int\ e^(3x)\ dx-int\ 4x\ dx# The left one can be solved using a u-substitution with #u=3x# and #(du)/dx=3#: #int\ e^(3x)\ dx=int\ e^u/3\ du=e^u/3=e^(3x)/3# So now we know: #int\ e^(3x)-4x\ dx=e^(3x)/3-2x^2# We can then evaluate the definite integral: #int_1^2e^(3x)-4x\ dx=[e^(3x)/3-2x^2]_1^2=(1/3e^6-8)-(1/3e^3-2)# #=1/3e^6-1/3e^3-8+2=1/3e^6-1/3e^3-6~=121.78#

Cameron Alexander · Answer

undefined To find the net area between $ f(x) = e^{3x} - 4x $ and the x-axis over the interval $[1, 2]$, you need to calculate the definite integral of $ f(x) $ from 1 to 2 and then take the absolute value since the function dips below the x-axis.

$ \text{Net area} = \left| \int_{1}^{2} (e^{3x} - 4x) \, dx \right| $

Now, integrate $ e^{3x} $ and $ 4x $ separately within the given interval:

$ \int e^{3x} \, dx = \frac{e^{3x}}{3} + C $

$ \int 4x \, dx = 2x^2 + C $

Then evaluate these integrals at the upper and lower limits:

$ \left. \left( \frac{e^{3x}}{3} - 2x^2 \right) \right|_{1}^{2} $

Plug in the upper and lower limits:

$ \left( \frac{e^{3(2)}}{3} - 2(2)^2 \right) - \left( \frac{e^{3(1)}}{3} - 2(1)^2 \right) $

$ = \left( \frac{e^6}{3} - 8 \right) - \left( \frac{e^3}{3} - 2 \right) $

$ = \frac{e^6}{3} - 8 - \frac{e^3}{3} + 2 $

$ = \frac{e^6 - e^3}{3} - 6 $

Finally, take the absolute value:

$ \text{Net area} = \left| \frac{e^6 - e^3}{3} - 6 \right| $