The length of a rectangle is 3 meters less than twice the width. if the area of the rectangle is 527 square meters, find the di
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Using the shell method, the volume integral would be
That is, each shell has a radius of <em>x</em> (the distance from a given <em>x</em> in the interval [0, 2] to the axis of revolution, <em>x</em> = 0) and a height equal to the difference between the boundary curves <em>y</em> = <em>x</em> ⁸ and <em>y</em> = 256. Each shell contributes an infinitesimal volume of 2<em>π</em> (radius) (height) (thickness), so the total volume of the overall solid would be obtained by integrating over [0, 2].
The volume itself would be
Using the disk method, the integral for volume would be
where each disk would have a radius of <em>x</em> = ⁸√<em>y</em> (which comes from solving <em>y</em> = <em>x</em> ⁸ for <em>x</em>) and an infinitesimal height, such that each disk contributes an infinitesimal volume of <em>π</em> (radius)² (height). You would end up with the same volume, 4096<em>π</em>/5.
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
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