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

Answer 1

#14 ln 2-3 e^4+e^2-3=-149.7# areal units, nearly. The negative integral indicates that area below the x-axis is quite large. Really, f < 0 and x > 8, and the graph is below the x-axis

#y = f(x) = x(ln x-e^x)#

The area # = int y dx#, with x from 2 to 4.

#= intx(ln x - e^x) dx#, with x from 2 to 4

#= int (ln xd(x^2/2)-xd(e^x))#, with x from 2 to 4

#=[x^2/2 ln x-int x^2/2 d(ln x) -(xe^x-int e^x dx)]#, x from 2 to 4.

#=[x^2/2 ln x -1/2 int x dx-xe^x+e^x]#, with x from 2 to 4

#=[1/4x^2(2 ln x -1 )-e^x(x-1)]@, between the limits x = 2 and 4.

Upon substitution of limits and simplification, this

#=14 ln 2-3 -3e^4+e^2#.

The graph inserted is suitably scaled to revveal the relevant portion of the curve

graph{x(ln x-e^x) [0, 4, -300, -0]}

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Answer 2

Adam Adkins

To find the net area between the function $f(x) = x\ln(x) - xe^x$ and the x-axis over $x$ in the interval $[2, 4]$ , you need to compute the definite integral of $f(x)$ from 2 to 4. The net area is the absolute value of this integral.

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Answer from HIX Tutor

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.