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Find the area of the region bounded by the given curves.

$ y = x^2 \ln x $ , $ y = 4 \ln x $

$\frac{16}{3} \ln (2)-\frac{29}{9}$

Integration Techniques

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Missouri State University

Campbell University

Baylor University

University of Michigan - Ann Arbor

the problem is find the area of the region only by the given curves here. Why is they go to High Square l N X and why is they go to four times l n X? First we lied. These two functions uh, echo and find intersection Point. Let's say this is X choir minus or times Helen likes is equal to zero. This X is greater than zero. We have X is equal to tool or Alan. Tax is 0 to 0. So this actually why Mhm So the interval is from 1 to 2 and what acts is between one and two. Four times L and X is great as on I squared times our next since I rear bounded by given curves is integral from 1 to 2 or times are impacts minus X choir times out next thanks, or this is equal to the integral from 1 to 2 four minus X choir. Alan Nex Next. Now we can use integration by parts to southeast. Integral former is integral of U K. Prime Detox is legal to you'll be minus the integral of your prime rejects. Now we can let you is the co two warm minus X choir. And we now when u is equal to Highland Max and we promise to four minus x squared, then well, you promise people wanna racks V a secretary or acts minus what third tax to three is power, then this integral. It's a cultural new times we So this is or X minus one third x two threes power times l and X from 12 to minus interior from 1 to 2. You prime time, sweetie. So this is more minus my dude X square. Mm x such as is you, can't you? So for the first term, we plug in two and 12 blocks than behalf. This is eight minus it over. Three times Ellen Chew and minus zero and minus. And to grow this oneness war axe minus one. Number nine x two threes. Power from one to this is true. 16/3 hour in two minus This one who is eight minus 8/9 minus four, minus one over I. So this is the result