Estimate the integral $\int_{1}^{10} \sqrt{x} d x$ using a left-hand sum and a right-hand sum with $n=3$ subdivisions.

NOTE: Enter exact answers or round to two decimal places.

Left-hand sum $=$

Right-hand sum $=$
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Step 1 ：First, we need to find the width of each subdivision. The interval from 1 to 10 is 9 units wide, so each of the 3 subdivisions will be 3 units wide.

Step 2 ：The left-hand sum is found by taking the height of the function at the left end of each subdivision and multiplying by the width of the subdivision. The heights are \(\sqrt{1}\), \(\sqrt{4}\), and \(\sqrt{7}\), so the left-hand sum is \(3(\sqrt{1} + \sqrt{4} + \sqrt{7}) = 3(1 + 2 + \sqrt{7}) = 9 + 3\sqrt{7}\).

Step 3 ：The right-hand sum is found by taking the height of the function at the right end of each subdivision and multiplying by the width of the subdivision. The heights are \(\sqrt{4}\), \(\sqrt{7}\), and \(\sqrt{10}\), so the right-hand sum is \(3(\sqrt{4} + \sqrt{7} + \sqrt{10}) = 3(2 + \sqrt{7} + \sqrt{10}) = 6 + 3\sqrt{7} + 3\sqrt{10}\).

Step 4 ：So, the left-hand sum is \(9 + 3\sqrt{7}\) and the right-hand sum is \(6 + 3\sqrt{7} + 3\sqrt{10}\).

Step 5 ：If we round to two decimal places, the left-hand sum is approximately \(\boxed{15.82}\) and the right-hand sum is approximately \(\boxed{21.48}\).