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

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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| )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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