적분법 전체 문제

Since $f(x)$ is linear, there exist real numbers $m$ and $c$ such that $f(x)=mx+c$. Now we can do some calculations. Suppose we have a rectangle in our Riemann sum that takes up the interval $[x,x+w]$.

• If we are using a left Riemann sum, our rectangle has height $f(x)=mx+c$. Then it has area $w(mx+c)$.
• If we are using a right Riemann sum, our rectangle has height $f(x+w)=m(x+w)+c=mx+c+mw$. Then it has area $w(mx+c+mw)$.
• If we are using a midpoint Riemann sum, our rectangle has height $f(x+\frac{1}{2}w)=m(x+\frac{1}{2}w)+c=mx+c+\frac{1}{2}mw$. Then it has area $w\left(mx+c+\frac{1}{2}w\right)$.

So, for each rectangle in our sums, the midpoint rectangle has the same area as the average of the left and right rectangles: $$w\left(mx+c+\frac{1}{2}mw\right) = \dfrac{ w(mx+c)+ w(mx+c+mw)}{2}$$ It follows that the midpoint Riemann sum has a value equal to the average of the values of the left and right Riemann sums. To see this, let the rectangles in the midpoint Riemann sum have areas $M_1,M_2,\ldots,M_n$, let the rectangles in the left Riemann sum have areas $L_1,L_2,\ldots,L_n$, and let the rectangles in the right Riemann sum have areas $R_1,R_2,\ldots,R_n$. Then the midpoint Riemann sum evaluates to $M_1+M_2+\cdots+M_n$, and:

$$\begin{aligned}\dfrac{ [L_1+L_2+\ldots+L_n]+ [R_1+R_2+\ldots+R_n] }{2} &= \dfrac{ L_1+ R_1 }{2}+ \dfrac{ L_2+ R_2 }{2}+ \cdots + \dfrac{ L_n+ R_n }{2} \\ &=M_1+M_2+\cdots+M_n\end{aligned}$$

So, the statement is true.

(Note, however, it is false for many non-linear functions $f(x)$.)