When estimating the area under a curve using left-bound rectangles, I know there will be an underestimate in a given interval if the function is increasing, $f'\left(x\right)>0$, and an overestimate if the function is decreasing, $f'\left(x\right)<0$.
Similarly, when using trapezoids, there will be an overestimate in intervals of upwards concavity, $f''\left(x\right)>0$, and an underestimate in intervals of downwards concavity, $f''\left(x\right)<0$.
Consider $f(x)=sin(x)+2$. I want to estimate the area $\int_{0}^{8}f\left(x\right)dx=17.1455$.
Using left-bound rectangles of width 1, my approximation is $\sum_{n=0}^{7}1f\left(n\right)=16.5537$. Could this "overall" underestimate be explained by the fact that $\int_{0}^{8}f'\left(x\right)dx>0$, that is to say, the "overall" gradient of the function within this interval is positive?
Similarly, using trapezoids of width 1, my approximation is $\sum_{n=0}^{7}\frac{1}{2}\left(f\left(n\right)+f\left(n+1\right)\right)=17.0484$. Again, could this "overall" underestimate be explained by the fact that $\int_{0}^{8}f''\left(x\right)dx<0$, that is to say, the "overall" concavity in this interval is downwards?
Thanks for your help :)