Question

If $f$ and $g$ are Riemann integrable, then $fg$ is Riemann Integrable (with a twist).

score 869 · Answer 1 · 2026-02-22 21:09:48

869

Views

If $f$ and $g$ are Riemann integrable, then $fg$ is Riemann Integrable (with a twist).

Published on 22 Feb 2026 - 21:09

score 838 · Answer 2 · 2015-11-09 19:19:17

838

Views

Prove that $\lim_{x\rightarrow \infty} \frac{1}{x}\int_0^x f(t)dt=c \mbox{.}$

Published on 09 Nov 2015 - 19:19

score 79 · Answer 3 · 2026-03-30 01:14:51

79

Views

Riemann integration of $\int_{[0,a]} f + \int_{[0,f(a)]} f^{-1}$

Published on 30 Mar 2026 - 1:14

score 101 · Answer 4 · 2015-11-11 18:46:00

101

Views

$F(x) = \int_x^{x+1} \sqrt {\arctan {t}}dt$ is bounded for $x \ge 0$.

Published on 11 Nov 2015 - 18:46

score 925 · Answer 5 · 2015-11-11 19:39:26

925

Views

Prove any function $f$ is Riemann integrable if it is bounded and continuous except finite number of points

Published on 11 Nov 2015 - 19:39

score 134 · Answer 6 · 2015-11-12 07:42:11

134

Views

Find $\lim_{x\to 0} \int_x^{x+1} \sqrt {\arctan {t}}\space dt$

Published on 12 Nov 2015 - 7:42

score 482 · Answer 7 · 2026-03-26 01:26:14

482

Views

Integration of subdifferential.

Published on 26 Mar 2026 - 1:26

score 112 · Answer 8 · 2015-11-13 11:06:13

Consider the sequence of functions defined as $ g_n =\int_{0}^{1} x^n f(x)dx$. Show that $ \frac {g_{n+1}}{g_n} \ge \frac {g_1}{g_0}$

score 228 · Answer 9 · 2015-11-13 17:24:20

228

Views

$f:[a,\infty)\to\mathbb R$ be continuous such that $\int_a^\infty f(x)dx$ exists finitely , then $\lim_{x \to \infty}f(x)=0$ ?

Published on 13 Nov 2015 - 17:24

score 115 · Answer 10 · 2026-04-01 23:06:50

115

Views

Product of improper integrable riemann function and integrable function.

Published on 01 Apr 2026 - 23:06

score 325 · Answer 11 · 2015-11-15 08:19:37

325

Views

$f\in \mathcal R[0,1]$ , $g \in C(\mathbb R)$ with period $1$ , then $\lim_{n \to \infty}\int_0 ^1 f(x)g(nx)dx=\int_0^1f(x)dx\int_0^1 g(x)dx$?

Published on 15 Nov 2015 - 8:19

score 270 · Answer 12 · 2026-03-30 03:05:09

Given $f(x)$, a continuous function on [0, 1] st $f(x)≥0$ for all $x∈[0, 1]$, show that if $\int_0^1 f(x)dx=0$ then $f(x) = 0$ for all $x ∈ [0, 1]$

score 1.3k · Answer 13 · 2026-03-27 06:21:40

1.3k

Views

Justification for arc length formula using Riemann Sum

Published on 27 Mar 2026 - 6:21

score 599 · Answer 14 · 2015-11-17 11:40:05

599

Views

Proof that ${\bf 1}_{\mathbb{Q}}(x) $ (indicator function of the set of rational numbers) is not Riemann-Integrable over [0,1]

Published on 17 Nov 2015 - 11:40

score 334 · Answer 15 · 2015-11-18 09:50:51

334

Views

Why is $\int_{0}^{1}\frac{1}{x} dx = \int_{1}^{+\infty}\frac{1}{x} dx = +\infty$?

Published on 18 Nov 2015 - 9:50

List Question

If $f$ and $g$ are Riemann integrable, then $fg$ is Riemann Integrable (with a twist).

Prove that $\lim_{x\rightarrow \infty} \frac{1}{x}\int_0^x f(t)dt=c \mbox{.}$

Riemann integration of $\int_{[0,a]} f + \int_{[0,f(a)]} f^{-1}$

$F(x) = \int_x^{x+1} \sqrt {\arctan {t}}dt$ is bounded for $x \ge 0$.

Prove any function $f$ is Riemann integrable if it is bounded and continuous except finite number of points

Find $\lim_{x\to 0} \int_x^{x+1} \sqrt {\arctan {t}}\space dt$

Integration of subdifferential.

Consider the sequence of functions defined as $ g_n =\int_{0}^{1} x^n f(x)dx$. Show that $ \frac {g_{n+1}}{g_n} \ge \frac {g_1}{g_0}$

$f:[a,\infty)\to\mathbb R$ be continuous such that $\int_a^\infty f(x)dx$ exists finitely , then $\lim_{x \to \infty}f(x)=0$ ?

Product of improper integrable riemann function and integrable function.

$f\in \mathcal R[0,1]$ , $g \in C(\mathbb R)$ with period $1$ , then $\lim_{n \to \infty}\int_0 ^1 f(x)g(nx)dx=\int_0^1f(x)dx\int_0^1 g(x)dx$?

Given $f(x)$, a continuous function on [0, 1] st $f(x)≥0$ for all $x∈[0, 1]$, show that if $\int_0^1 f(x)dx=0$ then $f(x) = 0$ for all $x ∈ [0, 1]$

Justification for arc length formula using Riemann Sum

Proof that ${\bf 1}_{\mathbb{Q}}(x) $ (indicator function of the set of rational numbers) is not Riemann-Integrable over [0,1]

Why is $\int_{0}^{1}\frac{1}{x} dx = \int_{1}^{+\infty}\frac{1}{x} dx = +\infty$?

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