[ L_n = \frac2n [4n + 3(n-1)] = \frac2n (7n - 3) = 14 - \frac6n ]
: [ R_4 = 0.5 [f(0.5) + f(1) + f(1.5) + f(2)] = 0.5 [0.25 + 1 + 2.25 + 4] = 0.5 \times 7.5 = 3.75 ]
Since I cannot directly generate or send a PDF file, this guide provides the , step-by-step solved exercises , and recommendations for you to copy into a document and save as PDF. 📘 Guide: Riemann Sums – Theory & Solved Exercises (PDF format) 1. Theoretical Summary Riemann Sum – approximates the definite integral (\int_a^b f(x) , dx):
[ M_4 \approx \frac\pi2 \times 1.306563 \approx 1.896 ] sumas de riemann ejercicios resueltos pdf
Similarly, (R_n = 14 + \frac6n) (check: (R_n = L_n + \Delta x (f(b)-f(a)))? (f(b)-f(a)=6,\ \Delta x \cdot 6 = \frac12n), but careful – compute:)
Note: (\sin(5\pi/8) = \sin(3\pi/8),\ \sin(7\pi/8) = \sin(\pi/8))
Sum: (\sum_i=0^n-1 4 = 4n,\ \sum_i=0^n-1 \frac6in = \frac6n \cdot \fracn(n-1)2 = 3(n-1)) [ L_n = \frac2n [4n + 3(n-1)] =
[ \int_a^b f(x) , dx = \lim_n \to \infty \sum_i=1^n f(x_i^*) \Delta x ]
: (\int_0^2 x^2 dx = \fracx^33 \Big|_0^2 = \frac83 \approx 2.6667)
Exact: (\int_0^\pi \sin x , dx = 2). So (M_4 \approx 1.896) (error (\approx 0.104)). Express (\lim_n \to \infty \frac1n \sum_i=1^n \left(1 + \fracin\right)^3) as an integral. (f(b)-f(a)=6,\ \Delta x \cdot 6 = \frac12n), but
Numerically: (\sin(22.5^\circ) \approx 0.382683,\ \sin(67.5^\circ) \approx 0.923880), sum (\approx 1.306563)
So: [ M_4 = \frac\pi4 \left[ 2\sin(\pi/8) + 2\sin(3\pi/8) \right] = \frac\pi2 [\sin(22.5^\circ) + \sin(67.5^\circ)] ]
Exact: (\int_1^3 (3x+1)dx = \left[\frac3x^22 + x\right]_1^3 = \left(\frac272+3\right) - \left(\frac32+1\right) = (13.5+3)-(1.5+1)=16.5-2.5=14)
Better: (R_n = \frac2n \sum_i=1^n (4 + \frac6in) = \frac2n[4n + \frac6n\cdot \fracn(n+1)2] = \frac2n[4n + 3(n+1)] = 14 + \frac6n)