The diagram shows part of the curve $y = \frac{1}{x^2}$.

1. Set up the definite integral for the area

$\text{Area} = \int_{1}^{6} \frac{1}{x^2} \, dx$

2. Integrate the function

$\int_{1}^{6} x^{-2} \, dx = \left[ \frac{x^{-1}}{-1} \right]_{1}^{6} = \left[ -\frac{1}{x} \right]_{1}^{6}$

3. Evaluate the definite integral

$\left( -\frac{1}{6} \right) - \left( -\frac{1}{1} \right) = -\frac{1}{6} + 1 = \frac{5}{6}$

1. Calculate the area under the curve from part (i)

Area $= \int_{1}^{6} x^{-2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{6} = -\frac{1}{6} - (-1) = \frac{5}{6}$

2. Relate the given sum to the area of rectangles under the curve

The sum $\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2}$ represents the total area of 5 rectangles, each of width 1, with heights corresponding to the $y$-values of the curve at $x = 2, 3, 4, 5, 6$.
Since the curve $y = \frac{1}{x^2}$ is decreasing, these rectangles lie entirely below the curve for the interval $x=1$ to $x=6$.

3. Conclude the inequality

Therefore, the sum of the areas of these rectangles is strictly less than the area under the curve from $x=1$ to $x=6$.
$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} < \int_{1}^{6} \frac{1}{x^2} \, dx$
$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} < \frac{5}{6}$