Find the area of the region between y=3e^(-2x) and y=lnx for 1<x<2. Round your answer to three decimal places.
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Originally Posted by titansfreak93 Find the area of the region between y=3e^(-2x) and y=lnx for 1<x<2. Round your answer to three decimal places. $\displaystyle 3e^{-2x} = \ln{x}$ at $\displaystyle x \approx 1.268$ $\displaystyle A = \int_1^{1.268} 3e^{-3x} - \ln{x} \, dx + \int_{1.268}^2 \ln{x} - 3e^{-2x} \, dx $ get out your calculator
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