selective-intregals-

Question Number 120628 by TANMAY PANACEA last updated on 01/Nov/20

s e l e c t i v e i n t r e g a l s

Commented by TANMAY PANACEA last updated on 01/Nov/20

t h a n k s

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

{\int_{- \frac{π}{2}}}^{\frac{π}{2}} \frac{d x}{e^{s i n x} + 1} . p = I

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d x}{e^{- s i n x} + 1} = I

2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} d x

I = \frac{π}{2}

Commented by mathmax by abdo last updated on 01/Nov/20

I = \int_{0}^{\frac{π}{3}} [\sqrt{3} tanx] dx changement \sqrt{3} tanx = t give tanx = \frac{t}{\sqrt{3}} \Rightarrow x = \arctan (\frac{t}{\sqrt{3}})

I = \int_{0}^{3} [t] \times \frac{1}{\sqrt{3} (1 + \frac{t^{2}}{3})} dt = \sqrt{3} \int_{0}^{3} \frac{[t]}{t^{2} + 3} dt =

= \sqrt{3} {\int_{0}^{1} 0 dt + \int_{1}^{2} \frac{1}{t^{2} + 3} dt + \int_{2}^{3} \frac{2}{t^{2} + 3} dt}

\int_{1}^{2} \frac{dt}{t^{2} + 3} =_{t = \sqrt{3} u} \int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}} \frac{\sqrt{3} du}{3 (1 + u^{2})} = \frac{\sqrt{3}}{3} {[arctanu]}_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}

= \frac{\sqrt{3}}{3} {\arctan (\frac{2}{\sqrt{3}}) - \arctan (\frac{1}{\sqrt{3}})}

\int_{2}^{3} \frac{2 dt}{t^{2} + 3} =_{t = \sqrt{3} u} \int_{\frac{2}{\sqrt{3}}}^{\sqrt{3}} \frac{2 \sqrt{3} du}{3 (1 + u^{2})} = \frac{2 \sqrt{3}}{3} {\arctan (\sqrt{3}) - \arctan (\frac{2}{\sqrt{3}})} \Rightarrow

I = \arctan (\frac{2}{\sqrt{3}}) - \arctan (\frac{1}{\sqrt{3}}) + 2 \arctan (\sqrt{3}) - 2 \arctan (\frac{2}{\sqrt{3}})

= - \arctan (\frac{2}{\sqrt{3}}) - (\frac{π}{2} - \arctan (\sqrt{3})) + 2 \arctan (\sqrt{3})

= 3 \arctan (\sqrt{3}) - \frac{π}{2} - \arctan (\frac{2}{\sqrt{3}})

we have \arctan (\sqrt{3}) = \frac{π}{3} \Rightarrow I = π - \frac{π}{2} - \arctan (\frac{2}{\sqrt{3}})

I = \frac{π}{2} - \arctan (\frac{2}{\sqrt{3}}) (\to c)

Commented by mathmax by abdo last updated on 01/Nov/20

A = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{dx}{e^{sinx} + 1} we do the changement x = - t \Rightarrow

A = - \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{- dt}{e^{- sint} + 1} = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{dx}{e^{- sinx} + 1} = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{e^{sinx}}{1 + e^{sinx}} dx \Rightarrow

2 A = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{1 + e^{sinx}} + \frac{e^{sinx}}{1 + e^{sinx}}) dx = \int_{- \frac{π}{2}}^{\frac{π}{2}} dx = π \Rightarrow A = \frac{π}{2}

Commented by mathmax by abdo last updated on 01/Nov/20

\lim_{x \to 1^{+}} \frac{\int_{1}^{x} ∣ t - 1 ∣ dt}{\sin (x - 1)} =_{t - 1 = u} \lim_{x \to 1^{+}} \frac{\int_{0}^{x - 1} ∣ u ∣ du}{\sin (x - 1)}

= \lim_{x \to 1^{+}} \frac{f^{^{'}} (x)}{g^{^{'}} (x)} = \lim_{x \to 1^{+}} \frac{x - 1}{(x - 1) \cos (x - 1)} = \lim_{x \to 1^{+}} \frac{1}{\cos (x - 1)}

= 1

Commented by TANMAY PANACEA last updated on 01/Nov/20

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 01/Nov/20

let f (x) = \frac{1}{n^{3}} {[1^{2} x] + [2^{2} x] + \dots . . [n^{2} x]} = \frac{1}{n^{3}} \sum_{k = 1}^{n} [k^{2} x] we have

[u] ⩽ u < [u] + 1 \Rightarrow u - 1 < [u] ⩽ u \Rightarrow k^{2} x - 1 < [k^{2} x] ⩽ k^{2} x \Rightarrow

\sum_{k = 1}^{n} (k^{2} x - 1) < \sum_{k = 1}^{n} [k^{2} x] ⩽ \sum_{k = 1}^{n} k^{2} x \Rightarrow

x \sum_{k = 1}^{n} k^{2} - n < \sum_{k = 1}^{n} [k^{2} x] ⩽ x \sum_{k = 1}^{n} k^{2} \Rightarrow

\frac{x}{6} n (n + 1) (2 n + 1) - n < \sum_{k = 1}^{n} [k^{2} x] ⩽ \frac{xn (n + 1) (2 n + 1)}{6} \Rightarrow

\frac{xn (n + 1) (2 n + 1)}{{6 n}^{3}} - \frac{1}{n^{2}} < f (x) ⩽ \frac{xn (n + 1) (2 n + 1)}{6} we have

\lim_{n \to + \infty} \frac{x}{{6 n}^{3}} n (n + 1) (2 n + 1) = {xlim}_{n \to + \infty} \frac{{2 n}^{3}}{{6 n}^{3}} = \frac{x}{3} \Rightarrow

\lim_{n \to + \infty} f (x) = \frac{x}{3}

Commented by mathmax by abdo last updated on 01/Nov/20

Q 45 is not clear

Commented by TANMAY PANACEA last updated on 01/Nov/20

g r e a t

Commented by TANMAY PANACEA last updated on 01/Nov/20

o k s i r l e t m e s e e t h e s o u r c e b o o k

Commented by Bird last updated on 01/Nov/20

y o u a r e w e l c o m e

Commented by bobhans last updated on 01/Nov/20

(42) \lim_{x \to 1} \frac{\underset{1}{\int^{x}} ∣ t - 1 ∣ dt}{\sin (x - 1)} = \lim_{x \to 1} \frac{∣ x - 1 ∣}{\cos (x - 1)} = 0

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