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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).

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01:47

Frank Lin

00:31

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Campbell University

Harvey Mudd College

Baylor University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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The boundaries of the shad…

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point)The boundaries o…

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Find the area of the regio…

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Sketch the region enclosed…

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.

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