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The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the area of the region bounded by the given curves and lines. $$ y = x, y = \frac{1}{ \sqrt{x} }, x = 2 $$
Answer:
Let $A$ be the area we seek. The first step in the process is find where the first two curves interset. \begin{align*} y &= x \\ y &= \frac{1}{ \sqrt{x} } \\ x &= \frac{1}{ \sqrt{x} } \\ x^2 &= \frac{1}{x} \\ x^3 &= 1 \\ x &= \pm 1 \\ \end{align*} \begin{align*} A &= \int_1^2 x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} \Big|_1^2 \\ A &= 2(2)^{\frac{1}{2}} - 2 = 2(\sqrt{2} - 1) \\ \end{align*} The book's answer is $\frac{7 - 4\sqrt{2}}{2}$. Where did I go wrong?
One thought on what I might have done wrong. It appears to me, the problem only gives three sides to the region you are trying to find the area of. I did not include a graph in the solution but I did produce a graph on paper.

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    $\begingroup$ The area between the curves means that the integrand is $x-\frac{1}{\sqrt{x}}$ $\endgroup$ – Paul Aug 14 at 19:23
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You have the bounds correct, but since the area is bounded above by $x$ and below by $\frac{1}{\sqrt{x}}$ the integral should be $$\int_1^2 (x-\frac{1}{\sqrt{x}})dx $$ $$=(\frac{1}{2}x^2-2\sqrt{x} )|^2_1$$ $$=2 - 2\sqrt{2} - (\frac{1}{2}-2)$$ $$=\frac{7}{2}-2\sqrt{2}$$

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