# Average Value of a Derivative Connecting to the Integral

I'm currently working on an essay about Riemann Sums and I have encountered a problem concerning the error in left and right Riemann sums. If we consider a function,$f$, over the interval $[x{_i}, x_{i+1}]$ we know that the left Riemann sum will estimate the area as $f(x{_i)}\times[x_{i+1} - x_{i}]$ and the right as $f(x_{i+1)}\times[x_{i+1} - x_{i}]$. Now, I know that if $f(x)$ is strictly increasing on the interval, the left will underestimate and the right will overestimate. However, could I extend this to say that if the average value of the derivative in the interval is positive, the left will underestimate and the right will overestimate? Would it be fair to say that for larger average values of $f'(x)$, there is a greater error in both methods?

• How do you define “average value of the derivative in the interval”? – José Carlos Santos Jul 5 '17 at 11:07
• $0<\int_{a}^{b}f'/(b-a)$ only means that $f(b)>f(a)$. What if $f((a+b)/2)$ is huge compared to $f(a)$ and $f(b)$? – Bettybel Jul 5 '17 at 11:26

As far as your first question is concerned, one can think of: $$f(x)=x^2,\ x\in[-0.9,1]$$ Then, $f'(x)=2x$ and the average value of $f'$ is: $$\frac{1}{1-(-0.9)}\int_{-0.9}^1f'(x)dx=\frac{1}{1.9}(1-0.9^2)=\frac{0.19}{1.9}=0.1>0$$ So, the average value of the derivative is strictly positive. Let now, $I=\int_{-0.9}^1f(x)dx$. We can calculate, easily that: $$I=0.09.333\dots=0.090\overline{3}$$. However, one can see that $f(-0.9)\times[1-(-0.9)]=0.81\times1.9=1.539>I$, so, the left Riemann Sums clearly does not underestimate $I$.
By assuming that the average value of $f'$ is positive over an interval $[a,b]$, one could think as follows: $$\frac{1}{b-a}\int_a^bf'(t)dt=\frac{f(b)-f(a)}{b-a}>0$$ From the Mean Value Theorem, there exist a $\xi\in(a,b)$, such as: $$f'(\xi)=\frac{f(b)-f(a)}{b-a}>0$$ and from that you can take it as far as you want, depending on the properties of $f$.
As far as your second question is concerned, it seems to me that if you take $g(x)=Mf(x)$, $x\in[-0,9,1]$, where $M$ is an arbitrary positive constant, as large as you like, then, you will get that the average value of $g'$ is: $$\frac{1}{1.9}\int_{-0.9}^1g'(x)dx=\frac{1}{1.9}M\int_{-0.9}^1f'(x)dx=\dots=0.1\cdot M$$ so, by letting $M\to+\infty$, it is evident that the average value of $g'$ can get as large as possible, however, the left Riemann Sum still overestimates the integral $\int\limits_{-0.9}^1g(x)dx$.
Lastly, it seems in an intuitive level, that the error gets larger as the mean value of the derivetive tends to infinity, where as error one can consider the following (this is for the left sum; for the right sum one can work in the same way): $$E_L(f,a,b)=\left|f(a)(b-a)-\int_a^bf(x)dx\right|$$ One can re-write the above - taking into consideration that $\int\limits_a^bdx=b-a$ - as: $$E_L(f,a,b)=\left|\int_a^bf(a)dx-\int_a^bf(x)dx\right|=\left|\int_a^b[f(a)-f(x)]dx\right|$$ By the triangular inequality for integrals, we take: $$E_L(f,a,b)=\left|\int_a^b[f(a)-f(x)]dx\right|\leq\int_a^b\left|f(a)-f(x)\right|dx\leq\int_a^b\max\limits_{x\in[a,b]}\{|f(a)-f(x)|\}dx=B(f,a,b)$$ So, we have found an upper bound - $B(f,a,b)$ - for our error. Let's assume that the average value of $f'$ is positive and equal to $N$, so: $$0<N=\frac{1}{b-a}\int_a^bf'(x)dx=\frac{f(b)-f(a)}{b-a}\Rightarrow f(b)-f(a)=N(b-a)$$ It is evident, by the definition of $B(f,a,b)$, that $B(f,a,b)\geq\int_a^b|f(b)-f(a)|dx$, so, we have: $$B(f,a,b)\geq\int_a^b|f(b)-f(a)|dx=\int_a^bN(b-a)dx=N(b-a)^2$$ So, letting $N\to+\infty$, gives that: $$B(f,a,b)\to+\infty$$ Which means that the error - well, actually our estimation of the error - increases infinitely, as the average value of $f'$ increases infinitely.