1)∫_1 ^3 (dx/(x^2 +[x]^2 +1−2x[x])) 2)∫_(−1) ^1 [x[1+sinπx]+1]dx 3)∫_0 ^2 x^([x^2 +1]) dx 4)∫_0 ^1 e^(2x−[2x]) d(x−[x]) 5)∫_(−2) ^3 ∣x^2 −5x−6∣dx from question 1 to 4 [.]←greatest integer fuction ∣.∣←mod questions taken from B.Stat entrance exam...


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Question Number 52619 by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$1) \int_{1}^{3} \frac{d x}{x^{2} + {[x]}^{2} + 1 - 2 x [x]}$ $2) \int_{- 1}^{1} [x [1 + s i n π x] + 1] d x$ $3) \int_{0}^{2} x^{[x^{2} + 1]} d x$ $4) \int_{0}^{1} e^{2 x - [2 x]} d (x - [x])$ $5) \int_{- 2}^{3} ∣ x^{2} - 5 x - 6 ∣ d x$ $f r o m q u e s t i o n 1 t o 4 [.] \leftarrow g r e a t e s t i n t e g e r f u c t i o n$ $∣ . ∣\leftarrow m o d$ $q u e s t i o n s t a k e n f r o m B . S t a t e n t r a n c e e x a m . . .$
Commented by maxmathsup by imad last updated on 10/Jan/19

$1) l e t I = \int_{1}^{3} \frac{d x}{x^{2} + {[x]}^{2} + 1 - 2 x [x]} \Rightarrow I = \int_{1}^{3} \frac{d x}{{(x - [x])}^{2} + 1}$ $= \int_{1}^{2} \frac{d x}{{(x - 1)}^{2} + 1} + \int_{2}^{3} \frac{d x}{{(x - 2)}^{2} + 1} b u t$ $\int_{1}^{2} \frac{d x}{{(x - 1)}^{2} + 1} =_{x - 1 = t} \int_{0}^{1} \frac{d t}{t^{2} + 1} = {[a r c t a n (t)]}_{0}^{1} = \frac{π}{4}$ $\int_{2}^{3} \frac{d x}{{(x - 2)}^{2} + 1} =_{x - 2 = t} \int_{0}^{1} \frac{d t}{t^{2} + 1} = \frac{π}{4} \Rightarrow I = \frac{π}{2} .$
Commented by maxmathsup by imad last updated on 10/Jan/19

$2) l e t A = \int_{- 1}^{1} [x [1 + s i n (π x)] + 1] d x \Rightarrow$ $A = \int_{- 1}^{1} [x + x [s i n (π x)] + 1] d x = \int_{- 1}^{1} [x + 1 + x [s i n (π x)]] d x$ $=_{x + 1 = t} \int_{0}^{2} [t + (t - 1) [- s i n (π t)] d t$ $= \int_{0}^{1} [t + (t - 1) [- s i n (π t)]] d t + \int_{1}^{2} [t + (t - 1) [- s i n (π t)] d t b u t$ $\int_{0}^{1} [t + (t - 1) [- s i n (π t)]] d t = \int_{0}^{\frac{1}{2}} [t + (t - 1) (0)] d t + \int_{\frac{1}{2}}^{1} [t + (t - 1) (0)] d t$ $= \int_{0}^{\frac{1}{2}} [t] d t + \int_{\frac{1}{2}}^{1} [t] d t = 0 + 0 = 0$ $\int_{1}^{2} [t + (t - 1) [- s i n (π t)]] d t = \int_{1}^{\frac{3}{2}} [t + (t - 1) (- 1)] d t + \int_{\frac{3}{2}}^{2} [t + (t - 1) (- 1)] d t$ $= \int_{1}^{\frac{3}{2}} [1] d t + \int_{\frac{3}{2}}^{2} [1] d t = \frac{3}{2} - 1 + 2 - \frac{3}{2} = 1 \Rightarrow A = 1 + {(p e r h a p s)}^{2} . . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$t h a n k y o u s i r . .$
Commented by Abdo msup. last updated on 10/Jan/19

$4) c h a n g e m e n t 2 x = t g i v e$ $\int_{0}^{1} e^{2 x - [2 x]} d (x - [x]) = \int_{0}^{2} e^{t - [t]} d (\frac{t}{2} - [\frac{t}{2}])$ $= \int_{0}^{1} e^{t - [t]} d (\frac{t}{2} - [\frac{t}{2}]) + \int_{1}^{2} e^{t - [t]} d (\frac{t}{2} - [\frac{t}{2}])$ $= \int_{0}^{1} e^{t} \frac{d t}{2} + \int_{1}^{2} e^{t - 1} \frac{d t}{2}$ $= \frac{1}{2} {[e^{t}]}_{0}^{1} + \frac{1}{2 e} {[e^{t}]}_{1}^{2} = \frac{1}{2} (e - 1) + \frac{1}{2 e} (e^{2} - 1) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$3) \int_{0}^{2} x^{[x^{2} + 1]} d x$ $\int_{0}^{1} x d x + \int_{1}^{\sqrt{2}} x^{2} d x + \int_{\sqrt{2}}^{\sqrt{3}} x^{3} d x + \int_{\sqrt{3}}^{2} x^{4} d x$ $n o w i t c a n b e s o l v e d . . . a t t a c h i n g g r a p h f o r c l a r i t y . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$f o r 1 > x ⩾ 0 [x^{2} + 1] = 1$ $\sqrt{2} > x ⩾ 1 [x^{2} + 1] = 2$ $\sqrt{3} > x ⩾ \sqrt{2} [x^{2} + 1] = 3$ $2 > x ⩾ \sqrt{3} [x^{2} + 1] = 4$
	Answered by ajfour last updated on 10/Jan/19

	$5) I = \int_{- 2}^{3} ∣ (x + 1) (x - 6) ∣ d x$ $l e t x + 2 = t$ $\Rightarrow I = \int_{0}^{3} ∣ (t - 1) (t - 8) ∣ d t$ $= (\frac{t^{3}}{3} - \frac{9 t^{2}}{2} + 8 t) ∣_{0}^{1} - (\frac{t^{3}}{3} - \frac{9 t^{2}}{2} + 8 t) ∣_{1}^{5}$ $= 2 (\frac{1}{3} - \frac{9}{2} + 8) - (\frac{125}{3} - \frac{225}{2} + 40)$ $= \frac{2}{3} - 9 + 16 - 41 - \frac{2}{3} + 112 + \frac{1}{2} - 40$ $= \frac{1}{2} + 128 - 90 = \frac{77}{2} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$t h a n k y o u s i r . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$4) \int_{0}^{1} e^{2 x - [2 x]} d (x - [x])$ $[2 x] = 0 \frac{1}{2} > x ⩾ 0$ $[x] = 0 \frac{1}{2} > x ⩾ 0$ $[2 x] = 1 1 > x ⩾ \frac{1}{2}$ $[x] = 0$ $\int_{0}^{\frac{1}{2}} e^{2 x - 0} d (x - 0) + \int_{\frac{1}{2}}^{1} e^{2 x - 1} d (x - 0)$ $= \int_{0}^{\frac{1}{2}} e^{2 x} d x + \int_{\frac{1}{2}}^{1} e^{2 x - 1} d x$ $n o w c a n b e s o l v e d . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

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