1. The problem is to find the sum of all integers from 1 up to 62. 2. We use the formula for the sum of the first $n$ natural numbers: $$S = \frac{n(n+1)}{2}$$ where $n=62$. 3. Substitute $n=62$ into the formula: $$S = \frac{62(62+1)}{2} = \frac{62 \times 63}{2}$$. 4. Calculate the numerator: $$62 \times 63 = 3906$$. 5. Now divide by 2: $$S = \frac{3906}{2}$$. 6. Show cancellation: $$S = \frac{\cancel{3906}}{\cancel{2}} = 1953$$. 7. Therefore, the sum of integers from 1 to 62 is $1953$.