Let $f$ be a decreasing function on $[0,1]$ with $f(0) = 7$ and $f(1) = 0$.
Show, using the definition of the Riemann sum, that the error in approximating $\int_{0}^{1}f(x)dx$ with a Riemann sum using right endpoints and n intervals of equal length is bounded above by $\frac{7}{n}$.
I don't seem to have any idea how to solve this question. Can someone please help me?
As you say the Riemann sum is $\sum_{k=0}^nf(\frac{k}{n})\frac{1}{n}$. Over the interval $(\frac{k-1}{n},\frac{k}{n})$ the function is decreasing and we picked the value at the right endpoint, which is the smallest value. So the biggest error in $f(x)$ for any point $x\in(\frac{k-1}{n},\frac{k}{n})$ is $f(\frac{k-1}{n})-f(\frac{k}{n})$. Hence the biggest possible error over that interval is $(f(\frac{k-1}{n})-f(\frac{k}{n}))\cdot\frac{1}{n}$.
Hence the max total error is $\frac{1}{n}\sum_{k=1}^n(f(\frac{k-1}{n})-f(\frac{k}{n}))$. The sum is $f(0)-f(\frac{1}{n})+f(\frac{1}{n})-f(\frac{2}{n})+\dots+f(\frac{n-1}{n})-f(1)$ which is just $f(0)-f(1)=7$. So the max total error is $\frac{7}{n}$.