# How do you find the area under the graph of #f(x)=e^(-2lnx)# on the interval [1, 2]?

Thus the area under the curve is

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To find the area under the graph of ( f(x) = e^{-2 \ln x} ) on the interval [1, 2], you need to integrate the function over that interval. First, rewrite the function in terms of ( x ) to simplify it: ( e^{-2 \ln x} = e^{\ln x^{-2}} = x^{-2} ). Now, integrate ( x^{-2} ) from 1 to 2 with respect to ( x ).

[ \int_{1}^{2} x^{-2} , dx = \left[ -x^{-1} \right]_{1}^{2} = -\frac{1}{2} + \frac{1}{1} = \frac{1}{2} ]

So, the area under the graph of ( f(x) = e^{-2 \ln x} ) on the interval [1, 2] is ( \frac{1}{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.

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