下图所示为 $y = \sqrt{x+40}$ 的图像。

1. Determine the area of the shaded rectangle.

The width of the shaded rectangle is $(n+1) - n = 1$.
Its height is the $y$-value of the curve at $x=n$, which is $\sqrt{n+40}$.
$\text{Area of shaded rectangle} = 1 \times \sqrt{n+40} = \sqrt{n+40}$

2. Compare the area of the rectangle to the area under the curve to establish the inequality.

The area under the curve $y = \sqrt{x+40}$ from $x=n$ to $x=n+1$ is given by the definite integral $\int_n^{n+1} \sqrt{x+40}\, dx$.
From the graph, since the curve is strictly increasing, the top edge of the shaded rectangle lies entirely under the curve.
Therefore, the area of the shaded rectangle is strictly less than the area under the curve:
$\sqrt{n+40} < \int_n^{n+1} \sqrt{x+40}\, dx$ (Shown)

1. Apply the given inequality for values of $n$ from $-40$ to $40$ and sum them up.

Substitute $n = -40, -39, -38, \dots, 40$ into the inequality $\sqrt{n+40} < \int_n^{n+1} \sqrt{x+40}\, dx$ and sum both sides:
$\sum_{n=-40}^{40} \sqrt{n+40} < \sum_{n=-40}^{40} \int_n^{n+1} \sqrt{x+40}\, dx$
$\sqrt{0} + \sqrt{1} + \dots + \sqrt{80} < \int_{-40}^{-39} \sqrt{x+40}\, dx + \int_{-39}^{-38} \sqrt{x+40}\, dx + \dots + \int_{40}^{41} \sqrt{x+40}\, dx$

2. Combine the integrals on the right-hand side and simplify the left-hand side.

Using the additive property of definite integrals over adjacent intervals, the right-hand side combines into a single integral:
$\sum_{n=-40}^{40} \int_n^{n+1} \sqrt{x+40}\, dx = \int_{-40}^{41} \sqrt{x+40}\, dx$
Since $\sqrt{0} = 0$, the left-hand side simplifies to:
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{80}$
Therefore, we have deduced that:
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{80} < \int_{-40}^{41} \sqrt{x+40}\, dx$

1. Determine the area of a rectangle that overestimates the area under the curve.

Consider a rectangle on the interval $[n, n+1]$ whose height is the function's value at the right endpoint, $x=n+1$.

The area of this rectangle is:
$\text{Area} = \text{width} \times \text{height}$
$\text{Area} = ((n+1) - n) \times \sqrt{(n+1)+40} = 1 \times \sqrt{n+41} = \sqrt{n+41}$

2. Compare the rectangle's area with the definite integral to establish the inequality.

The definite integral $\int_n^{n+1} \sqrt{x+40}\, dx$ represents the exact area under the curve from $x=n$ to $x=n+1$.

Since the curve $y = \sqrt{x+40}$ is strictly increasing, the area of the rectangle with height determined by the right endpoint is strictly greater than the area under the curve.

$\text{Area of rectangle} > \text{Area under curve}$
$\sqrt{n+41} > \int_n^{n+1} \sqrt{x+40}\, dx \text{ (shown)}$

1. Substitute integer values of $n$ from $-40$ to $40$ into the inequality from part [2]

From part [2], we have the inequality:
$\sqrt{n+41} > \int_n^{n+1} \sqrt{x+40}\, dx$

Substituting $n = -40, -39, \dots, 40$:
When $n = -40$: $\quad \sqrt{1} > \int_{-40}^{-39} \sqrt{x+40}\, dx$
When $n = -39$: $\quad \sqrt{2} > \int_{-39}^{-38} \sqrt{x+40}\, dx$
When $n = -38$: $\quad \sqrt{3} > \int_{-38}^{-37} \sqrt{x+40}\, dx$
$\vdots$
When $n = 40$: $\quad \sqrt{81} > \int_{40}^{41} \sqrt{x+40}\, dx$

2. Sum the inequalities and combine the definite integrals

Summing the left hand side and right hand side of all these inequalities:
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{81} > \int_{-40}^{-39} \sqrt{x+40}\, dx + \int_{-39}^{-38} \sqrt{x+40}\, dx + \dots + \int_{40}^{41} \sqrt{x+40}\, dx$

Combining the integrals over the adjacent intervals gives the single definite integral from the lower limit $-40$ to the upper limit $41$:
$\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{81} > \int_{-40}^{41} \sqrt{x+40}\, dx \quad \text{(shown)}$

1. Evaluate the definite integral $\int_{-40}^{41} \sqrt{x+40} \, dx$ to find the numerical bounds.

$\int_{-40}^{41} \sqrt{x+40} \, dx = \left[ \frac{2}{3}(x+40)^{\frac{3}{2}} \right]_{-40}^{41} = \frac{2}{3}(81)^{\frac{3}{2}} - 0 = \frac{2}{3}(729) = 486$

2. Substitute the evaluated integral into the inequalities derived in parts [1] and [3].

From part [1]:
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{80} < 486$

From part [3]:
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{81} > 486$

3. Isolate the sum up to $\sqrt{80}$ in the second inequality and combine the bounds to deduce $a$.

Rewrite the sum from part [3] by separating the last term:
$(\sqrt{1} + \sqrt{2} + \dots + \sqrt{80}) + \sqrt{81} > 486$
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{80} + 9 > 486$
$\sqrt{1} + \sqrt{2} + \dots + \sqrt{80} > 477$

Combining the lower and upper bounds gives:
$477 < \sqrt{1} + \sqrt{2} + \dots + \sqrt{80} < 486$

Factoring out 9 to match the required form:
$9(53) < \sqrt{1} + \sqrt{2} + \dots + \sqrt{80} < 9(54)$

Comparing this with $9a < \sqrt{1} + \sqrt{2} + \dots + \sqrt{80} < 9(a+1)$, we obtain $a = 53$.