Finding the area under a curve when the area is bounded by 3 curves.

Question

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.

Answer 1

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 $$\int_1^2 (x-\frac{1}{\sqrt{x}})dx $$ $$=(\frac{1}{2}x^2-2\sqrt{x} )|^2_1$$ $$=2 - 2\sqrt{2} - (\frac{1}{2}-2)$$ $$=\frac{7}{2}-2\sqrt{2}$$