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Essential Calculus: Early Transcendentals
Found in: Page 370
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the number \(b\) such that the line \(y = b\) divides the region bounded by the curves \(y = \) \({x^2}\) and \(y = 4\) into two regions with equal area.

The value of \(b\) is \(2.52\).

See the step by step solution

Step by Step Solution

The area under a curve

The area under a curve between two points is found out by doing a definite integral between the two points.

Find \(b\) such that the blue and red areas are equal.

The regions \(y = {x^2},y = 4\) are divided by \(y = b\)

Integrate along the \(y\)-axis, so for \(y = {x^2}\) use \(x = \sqrt y \), which is the right half of the parabola. The right/left halves are symmetric so ignore the left half.

The red region is the integral from \(y = 0\) to \(b\), the blue region is from \(y = b\) to 4

Set the two integral areas equal to each other to solve for \(b\)

The value of \(b\) is given by

\(\int_0^b {\sqrt y \,} dy = \int_b^4 {\sqrt y \,} dy\)

\[\left[{\frac{2}{3}{y^{\frac{3}{2}}}} \right]_0^b = \left[ {\frac{2}{3}{y^{\frac{3}{2}}}} \right]_b^4\]

\({b^{\frac{3}{2}}} - 0 = {4^{\frac{3}{2}}} - {b^{\frac{3}{2}}}\)

\(2{b^{\frac{3}{2}}} = 8\)

\({b^{\frac{3}{2}}} = 4\)

\(b = {4^{\frac{2}{3}}}\)

\( \approx 2.52\)

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