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Find the area of the region in the first quadrant bounded on theleft by the $y$ -axis, below by the curve $x=2 \sqrt{y}$ , above left by thecurve $x=(y-1)^{2},$ and above right by the line $x=3-y .$
$A=\int_{0}^{1} 2 \sqrt{y} d y+\int_{1}^{2} 3-y-(y-1)^{2} d y$
Calculus 1 / AB
Chapter 5
Integrals
Section 6
Definite Integral Substitutions and the Area Between Curves
Campbell University
Harvey Mudd College
University of Nottingham
Idaho State University
Lectures
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In mathematics, precalculu…
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In mathematics, a function…
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Find the area of the regio…
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06:34
Find the area of the follo…
So we want to find the area of this region that's pictured here. So we're given several functions of X in terms of why, and you see how maybe we can sort of separate this into two different regions that are bounded by two curves. So if I draw this horizontal line, X is equal to sorry. This horizontal line Y is equal to one we see below the line. The region is bounded to the left by just the Y axis that's X equals zero and to the right, by this curb, excess to square roots of why and above that our upper bound is three minus wise, equal X, and are lower bound his exes. Why minus one squared. So the area we can write down as being the integral first from 0 to 1 with respect to why of to square roots of why minus zero. Okay, because the Y axis is the graph of X equals zero. All right, so then yes, sir, than d y here. And then we would just have the integral from 1 to 2. Now what's our upper function is three minus y and then our lower function is this one red. Okay, so these two in a girl's add that together gives you the total area of the shaded regions. Now we won't evaluate thes because it should be straightforward. You just expand here on the right hand side. You're just integrating polynomial. So to find the anti derivatives, just reverse the power rule. If you're confused about how to actually integrate thes these two look back to some previous videos where we work through all the steps and full detail.
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