Answer to Question #117675 in Calculus for Lizwi

Question #117675
Find the area of the region enclosed between the graphs of y=ex and y=e−x and the line x=ln2.
1
Expert's answer
2020-05-22T18:26:40-0400

The graphs of y=exy=e^x and y=exy=e^{-x} intersect if x=0,x=0, therefore we should integrate (exex)(e^x-e^{-x}) on the interval of [0, ln 2]:

0ln2(exex)dx=(ex+ex)0ln2=(2+0.5)(1+1)=0.5.\int\limits_0^{\ln 2} (e^x-e^{-x})\,dx = (e^x + e^{-x})\Big|_{0}^{\ln 2} = (2+0.5) - (1+1) = 0.5.


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Comments

Assignment Expert
26.05.20, 00:23

Dear Lebina Moletsane, please use the panel for submitting new questions.

Lebina Moletsane
24.05.20, 20:35

Find the volume of the solid with cross-sectional area 21−4x2√ lying between x=0 and x=12

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