What is the area of the region in the first quadrant by the graph of #y=e^(x/2)# and the line x=2?
2e-2 = 3.4366, nearly. See the graph for this area.
The line x = 2 is parallel to y-axis.
the curve, x = 0, x-axis and x = 2. See the graph.
graph{(x-2ln y)(2-x)(x)=0[0 2] [0 2.72]}
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To find the area of the region in the first quadrant bounded by the graph of (y = e^{x/2}) and the line (x = 2), we need to compute the definite integral of the function (e^{x/2}) with respect to (x) from (x = 0) to (x = 2).
[ A = \int_{0}^{2} e^{x/2} dx ]
Using integration techniques, the antiderivative of (e^{x/2}) is (2e^{x/2}).
[ A = \left[2e^{x/2}\right]_{0}^{2} = 2e^{2/2} - 2e^{0/2} ] [ A = 2e - 2 ]
Therefore, the area of the region in the first quadrant bounded by the graph of (y = e^{x/2}) and the line (x = 2) is (2e - 2).
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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.
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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