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If $ w'(t) $ is the rate of growth of a child in …

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Problem 50 Easy Difficulty

The boundaries of the shaded region are the $ y $-axis, the line $ y = 1 $, and the curve $ y = \sqrt[4]{x} $. Find the area of this region by writing $ x $ as a function of $ y $ and integrating with respect to $ y $ (as in Exercise 49).

Name: the boundaries of the shaded region are the y axis the line y 1 and the curve y sqrt4x find the area
Uploaded: 2021-07-31
Duration: 2 min 18 s
Channel: Numerade
Description: The boundaries of the shaded region are the $ y $-axis, the line $ y = 1 $, and the curve $ y = \sqrt[4]{x} $. Find the area of this region by writing $ x $ as a function of $ y $ and integrating with respect to $ y $ (as in Exercise 49).

Answer

$\frac{1}{5}$

Grace H.

Catherine R.

Missouri State University

Heather Z.

Oregon State University

Michael J.

Idaho State University

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Video Transcript

so to find the function that we want to integrate, we need to write this curve that were given as X. As a function of Y. So to do this we can just raise both sides to the 4th power or the fourth. It's this time of the 4th so it just becomes X Equals Y to the 4th. So we have a function that we want to integrate now we just need to find the upper and lower bounds. Um So we're giving the boundaries the Y axis. So lower bound, the y axis Is represented by the equation x equals zero. So we have a lower bound of zero. Our upper bounds We're defined from the line why is equal to one and so have X as a function of why we need to find X. Such that see why excuse me to take our original equation. So the one y equals one. This curve Has the intersect why it was one. So in this case X has to be one. So there we have our lower bound of zero upper bound of one. And we integrate this function Y to the 4th. With respect to Y. Valuing this, we can get anti derivative of Y. to the fourth is why to the 5th five upper bound one or ground zero, So one of the 5th for five minus here to the fifth over five is just 1/5 minus zero. So we get the answer of 1/5.

Joseph R.

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