Discrete vs. Continuous

Calculus Level 3

∑ x = 1 n ( a x + b ) − ∫ 0 n ( a x + b ) d x = ? \large \sum_{x=1}^n (ax+b) - \int_0^n (ax+b) \, dx = ?

Hint: There is a nice geometric interpretation of this that doesn't require any calculus, as long as you understand that the definite integral represents the area between the x x -axis and the function between x = 0 x=0 and x = n x=n .

1 2 ( b − a ) \frac{1}{2}(b-a) 1 2 a \frac{1}{2}a 1 2 a n 2 \frac{1}{2}an^2 1 2 ( a + b ) \frac{1}{2}(a+b) 1 2 a n \frac{1}{2}an

Mark Lama by Mark Lama , 40, USA

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1 solution

Mark Lama
Sep 19, 2017

The vertical bars (including both the red and blue areas) represent the value of the sum of the finite arithmetic series ( a + b ) + ( 2 a + b ) + ⋯ + ( n a + b ) (a+b)+(2a+b)+\dots+(na+b) The blue area under the line y = a x + b y=ax+b represents the value of the definite integral. The area of the red triangles represents the difference between the value of the sum and the value of the definite integral. The area of each red triangle is 1 2 a \frac{1}{2}a . Together, the area of all the red triangles is 1 2 a n \frac{1}{2}an , and this is our answer. Notice that we didn't have to evaluate either the sum or the integral, although this could be easily done. The sum works out to a n ( n + 1 ) 2 + b n a\frac{n(n+1)}{2}+bn , or 1 2 a n 2 + ( 1 2 a + b ) n \frac{1}{2}an^2+\left( \frac{1}{2}a+b \right)n , and the integral is clearly the area of the the large blue triangle plus the area of the large blue rectangle, or 1 2 a n 2 + b n \frac{1}{2}an^2+bn .

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