Your friend manages to squeeze in one problem between her classes. She used her calculator to find $L_n$, $R_n$, $T_n$, and $M_n$ to approximate a definite integral and wrote down the results on a piece of paper. The results were:
0.3676, 0.3212, 0.3412, and 0.2748.
She didn't write down the problem nor did she record the value of $n$ she used. But she did draw the function (over the relevant interval) on a paper. The function looks like this:
She wants help in labeling the data she recorded as a left-hand sum, a right-hand sum, a midpoint sum, and a trapezoidal sum. Which is which? Explain your answer. $$\\$$
I don't know if this is correct, but this is what I have so far:
I know $L_n$ is the largest overestimate, $M_n$ is the second largest overestimate, $R_n$ is an underestimate, and $T_n$ the second underestimate.
Thus I got:
$L_n$ → 0.3676
$R_n$ → 0.2748
$T_n$ → 0.3412
$M_n$ → 0.3212
When a function is decreasing: $L_n$ is an overestimate and $R_n$ is an underestimate.
When a function is increasing: $L_n$ is an underestimate and $R_n$ is an overestimate.
When a function is concave down: $T_n$ is an underestimate and $M_n$ is an overestimate.
When a function is concave up: $T_n$ is an overestimate and $M_n$ is an underestimate. $$\\$$ In your case, your graph is decreasing and is concaving down so $L_n$ is an overestimate, $R_n$ is an underestimate, $T_n$ is an underestimate, and $M_n$ is an overestimate.
$L_n$ = 0.3676, $R_n$ = 0.2748, $M_n$ = 0.3412, and $T_n$ = 0.3212.