Riemann Integral Problems And Solutions Pdf Apr 2026

\subsection*Problem 8 Evaluate (\lim_n\to\infty \frac1n\sum_k=1^n \sin\left(\frack\pi2n\right)).

\subsection*Solution 8 Rewrite: (\frac1n\sum_k=1^n \sin\left(\frack\pi2n\right) = \frac1\pi/2 \cdot \frac\pi2n\sum_k=1^n \sin\left(\frack\pi2n\right))? Actually: Let (\Delta x = \frac\pi/2n = \frac\pi2n), then the sum is (\frac1n\sum \sin(k\Delta x) = \frac2\pi\cdot \frac\pi2n\sum \sin(k\Delta x))? Wait: (\frac1n = \frac2\pi\cdot \frac\pi2n). So: [ \lim_n\to\infty \frac1n\sum_k=1^n \sin\left(\frack\pi2n\right) = \lim_n\to\infty \frac2\pi\sum_k=1^n \sin\left(\frack\pi2n\right)\cdot\frac\pi2n = \frac2\pi\int_0^\pi/2 \sin x,dx = \frac2\pi[-\cos x]_0^\pi/2 = \frac2\pi(0+1) = \frac2\pi. ] riemann integral problems and solutions pdf

\subsection*Problem 7 Prove that if (f) is continuous on ([a,b]), then (\int_a^b f(x),dx = \lim_n\to\infty \fracb-an\sum_k=1^n f\left(a + k\fracb-an\right)). then (\int_a^b f(x)

\subsection*Solution 9 Since (f \ge 0), any lower sum (L(P,f) \ge 0). The integral is the supremum of lower sums, hence (\int_a^b f = \sup L(P,f) \ge 0). any lower sum (L(P

If f ≥ 0 integrable, prove ∫ f ≥ 0.