Find the area of the region bounded by the curves y = x², y = 1/x and x = 1/2 (see the graphical representive figure):

Concept:

The area between any two curves in a 2-D plane (xy - plane) can be easily obtained using integration technique as shown:

\(A=\mathop{\int }_{y=c}^{y=d}\mathop{\int }_{x=a}^{x=b}dxdy\)

Where ‘A’ is the area bounded between two straight lines as shown:

Calculation:

Given equations are \(y = {x^2}\) and \(y = \frac{1}{x}\)

The intersection point of these two curves will be (1, 1)

Let consider a vertical strip as shown in the figure, then the limits will be:

X varies from 0.5 to 1 and y varies from \(\frac{1}{x}to\;{x^2}\)

So, the area can be obtained as follows:

\(\Rightarrow Area=\mathop{\int }_{x=1/2}^{x=1}\mathop{\int }_{y={{x}^{2}}}^{y=1/x}dy~dx\)

\(=\mathop{\int }_{x=\frac{1}{2}}^{x=1}\left. \left( y \right) \right|_{{{x}^{2}}}^{1/x}.dx\)

\(=\mathop{\int }_{x=1/2}^{x=1}\left( \frac{1}{x}-{{x}^{2}} \right)dx\)

\(\Rightarrow \mathop{\int }_{1/2}^{1}\frac{1}{x}dx-\mathop{\int }_{1/2}^{1}{{x}^{2}}dx\)

\(=\left. \ln \left( x \right) \right|_{1/2}^{1}\left. -\frac{{{x}^{3}}}{3} \right|_{1/2}^{1}\)

\(=\ln \left( 2 \right)-\frac{1}{3}\left( 1-\frac{1}{8} \right)~\)

\(=\ln \left( 2 \right)-\frac{7}{24}\)