1) ∫_0 ^( 1) ((1−x^7 )^(1/4) −(1−x^4 )^(1/7) )dx = ? 2) If f(x)=x^3 +3x+4 then the value of ∫_(−1) ^( 1) f(x)dx + ∫_0 ^( 4) f^( −1) (x)dx = ? 3) ∫_(−π) ^( π) (1+cosx+cos2x+....+cos13x)(1+sinx+...+sin13x)dx=? 4) ∫_0 ^( 2) ((√(x^3 +1)) + (x^2 +2x)^(1/3) )dx =? (This time func. are not inverse of each other,right ?)


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Question Number 54826 by rahul 19 last updated on 13/Feb/19

$1) \int_{0}^{1} ({(1 - x^{7})}^{\frac{1}{4}} - {(1 - x^{4})}^{\frac{1}{7}}) d x = ?$ $2) I f f (x) = x^{3} + 3 x + 4 t h e n t h e v a l u e o f$ $\int_{- 1}^{1} f (x) d x + \int_{0}^{4} f^{- 1} (x) d x = ?$ $3) \int_{- π}^{π} (1 + c o s x + c o s 2 x + . . . . + c o s 13 x) (1 + s i n x + . . . + s i n 13 x) d x = ?$ $4) \int_{0}^{2} (\sqrt{x^{3} + 1} + {(x^{2} + 2 x)}^{\frac{1}{3}}) d x = ?$ $(T h i s t i m e f u n c . a r e n o t i n v e r s e o f e a c h$ $o t h e r, r i g h t ?)$
Commented by maxmathsup by imad last updated on 12/Feb/19

$2) w e h a v e \int_{- 1}^{1} f (x) d x = \int_{- 1}^{1} (x^{3} + 3 x + 4) d x = {[\frac{x^{4}}{4} + \frac{3}{2} x^{2} + 4 x]}_{- 1}^{1}$ $= \frac{1}{4} + \frac{3}{2} + 4 - \frac{1}{4} - \frac{3}{2} + 4 = 8 l e t f i n d f^{- 1} (x)$ $f (x) = y \Leftrightarrow x^{3} + 3 x + 4 = y \Leftrightarrow x^{3} + 3 x + 4 - y = 0 c h a n g e m e n t x = u + v g i v e$ $u^{3} + 3 u v (u + v) + v^{3} + 3 (u + v) + 4 - y = 0 \Rightarrow$ $u^{3} + v^{3} + 4 - y + (u + v) (3 u v + 3) = 0 \Rightarrow u^{3} + v^{3} = y - 4 a n d u v = - 1 \Rightarrow$ $s o u^{3} a n d v^{3} a r e s o l u t i o n o f t h e e q u a t i o n X^{2} - (y - 4) X - 1 = 0 \Rightarrow$ $Δ = {(y - 4)}^{2} + 4 > 0 \Rightarrow X_{1} = \frac{y - 4 + \sqrt{{(y - 4)}^{2} + 4}}{2} a n d$ $X_{2} = \frac{y - 4 - \sqrt{{(y - 4)}^{2}) 4}}{2} \Rightarrow x =^{3} \sqrt{X_{1}} + \sqrt{X_{2}}$ $\Rightarrow f^{- 1} (x) = {(\frac{x - 4 + \sqrt{{(x - 4)}^{2} + 4}}{2})}^{\frac{1}{3}} + {(\frac{x - 4 - \sqrt{{(x - 4)}^{2} + 4}}{2})}^{\frac{1}{3}} \Rightarrow$ $\int_{0}^{4} f^{- 1} (x) d x = \frac{1}{(^{3} \sqrt{2})} \int_{0}^{4} {(x - 4 + \sqrt{{(x - 4)}^{2} + 4})}^{\frac{1}{3}} d x + \frac{1}{(^{3} \sqrt{2})} \int_{0}^{4} {(x - 4 - \sqrt{{(x - 4)}^{2} + 4})}^{\frac{1}{3}} d x$ $l e t I = \int_{0}^{4} {(x - 4 + \sqrt{{(x - 4)}^{2} + 4})}^{\frac{1}{3}} d x c h a n g . x - 4 = 2 s h (t) g i v e$ $I = \int_{- a r g s h (2)}^{0} {(2 s h (t) + 2 c h (t))}^{\frac{1}{3}} 2 c h (t) d t$ $= 2 (^{3} \sqrt{2}) \int_{- l n (2 + \sqrt{5})}^{0} (s h (t) + c h {(t)}^{\frac{1}{3}} c h (t) d t . . . . b e c o n t i n u e d . . .$ $l e t A = \int_{0}^{4} f^{- 1} (x) d x c h a n g e m e n t f^{- 1} (x) = t g i v e x = f (t) \Rightarrow$ $A = \int_{f^{- 1} (0)}^{f^{- 1} (4)} t f^{^{'}} (t) d t = {[t f (t)]}_{f^{- 1} (0)}^{f^{- 1} (4)} - \int_{f^{- 1} (0)}^{f^{- 1} (4)} f (t) d t$ $b u t f^{- 1} (4) = 1 - 1 = 0 a n d f^{- 1} (0) = {(\frac{- 4 + \sqrt{20}}{2})}^{\frac{1}{3}} + {(\frac{- 4 - \sqrt{20}}{2})}^{\frac{1}{3}}$ $= {(- 2 + \sqrt{5})}^{\frac{1}{3}} - {(2 + \sqrt{5})}^{\frac{1}{3}} = α_{0} \Rightarrow$ $A = {[t f (t)]}_{α_{0}}^{0} - \int_{α_{0}}^{0} f (t) d t = - α_{0} f (α_{0}) + \int_{0}^{α_{0}} (t^{3} + 3 t + 4) d t$ $= - α_{0} f (α_{0}) + \frac{α_{0}^{3}}{3} + \frac{3}{2} α_{0}^{2} + 4 α_{0}$ $s o t h e v a l u e o f \int_{0}^{4} f^{- 1} (x) d x i s k n o w n .$
Commented by maxmathsup by imad last updated on 12/Feb/19

$i n g e n e r a l l e t f i n d \int_{a}^{b} f^{- 1} (x) d x c h a n g . f^{- 1} (x) = t g i v e x = f (t) \Rightarrow$ $\int_{a}^{b} f^{- 1} (x) d x = \int_{f^{- 1} (a)}^{f^{- 1} (b)} t f^{^{'}} (t) d t = {[t f (t)]}_{f^{- 1} (a)}^{f^{- 1} (b)} - \int_{f^{- 1} (a)}^{f^{- 1} (b)} f (t) d t$ $= {b f}^{- 1} (b) - {a f}^{- 1} (a) - \int_{f^{- 1} (a)}^{f^{- 1} (b)} f (t) d t .$
Commented by kaivan.ahmadi last updated on 12/Feb/19

$y = f^{- 1} (x) \Rightarrow f (y) = x \Rightarrow y^{3} + 3 y + 4 = x, dx = ({3 y}^{2} + 3) dy$ $if x = 0 \Rightarrow y = - 1$ $if x = 4 \Rightarrow y = 0 or - 3$ $\int_{0}^{4} f^{- 1} (x) dx = \int_{- 1}^{0} y ({3 y}^{2} + 3) dy = {[\frac{{3 y}^{4}}{4} + \frac{{3 y}^{2}}{2}]}_{- 1}^{0} = - (\frac{3}{4} + \frac{3}{2}) = - \frac{9}{4}$ $is it true ?$
Commented by maxmathsup by imad last updated on 12/Feb/19
$3) let simlify A(x)=(1+cosx +cos(2x)+...+cos(3x))(1+sinx +sin(2x)+...+sin(13x)) let S_n (x)=Σ_(k=0) ^n cos(kx) and W_n (x)=Σ_(k=0) ^n sin(kx) ⇒ S_n +iW_n =Σ_(k=0) ^n (e^(ix) )^k =((1−(e^(ix) )^(n+1) )/(1−e^(ix) )) = ((1−e^(i(n+1)x) )/(1−cosx −isinx)) =((1−cos(n+1)x −isin(n+1)x)/(1−cosx −isinx)) =((2sin^2 (((n+1)x)/2)−2i sin((((n+1)x)/2))cos((((n+1)x)/2)))/(2 sin^2 ((x/2))−2i sin((x/2))cos((x/2)))) =((−i sin((((n+1)x)/2)) e^(i(((n+1)x)/2)) )/(−isin((x/2)) e^((ix)/2) )) =((sin((((n+1)x)/2))e^(i((nx)/2)) )/(sin((x/2)))) =((sin((((n+1)x)/2)))/(sin((x/2)))){ cos(((nx)/2)) +isin(((nx)/2))} ⇒ S_n (x) = ((sin((((n+1)x)/2)) cos(((nx)/2)))/(sin((x/2)))) and W_n (x)=((sin((((n+1)x)/2))sin(((nx)/2)))/(sin((x/2)))) ⇒ 1+cosx +cos(2x)+...cos(13x)=((sin(7x)cos(((13x)/2)))/(sin((x/2)))) and 1+sinx +sin(2x)+...+sin(13x) =1+((sin(7x)sin(((13x)/2)))/(sin((x/2)))) ⇒ (...)×(...) =((sin(7x) cos(((13x)/2)))/(sin((x/2)))) +((sin^2 (7x)sin(13x))/(2sin^2 ((x/2)))) ⇒ ∫_(−π) ^π A(x)dx = ∫_(−π) ^π ((sin(7x)cos(((13x)/2)))/(sin((x/2))))dx +∫_(−π) ^π ((sin^2 (7x)sin(13x))/(2sin^2 ((x/2))))dx ∫_(−π) ^π ((sin(7x)cos(((13x)/2)))/(sin((x/2)))) dx =2 ∫_0 ^π ((sin(7x)cos(((13x)/2)))/(sin((x/2))))dx (even function) ∫_(−π) ^π ((sin^2 (7x)sin(13x))/(2sin^2 ((x/2))))dx =0 ( odd function) ...be continued....$
$3) l e t s i m l i f y A (x) = (1 + c o s x + c o s (2 x) + . . . + c o s (3 x)) (1 + s i n x + s i n (2 x) + . . . + s i n (13 x))$ $l e t S_{n} (x) = \sum_{k = 0}^{n} c o s (k x) a n d W_{n} (x) = \sum_{k = 0}^{n} s i n (k x) \Rightarrow S_{n} + {i W}_{n}$ $= \sum_{k = 0}^{n} {(e^{i x})}^{k} = \frac{1 - {(e^{i x})}^{n + 1}}{1 - e^{i x}} = \frac{1 - e^{i (n + 1) x}}{1 - c o s x - i s i n x}$ $= \frac{1 - c o s (n + 1) x - i s i n (n + 1) x}{1 - c o s x - i s i n x} = \frac{2 {s i n}^{2} \frac{(n + 1) x}{2} - 2 i s i n (\frac{(n + 1) x}{2}) c o s (\frac{(n + 1) x}{2})}{2 {s i n}^{2} (\frac{x}{2}) - 2 i s i n (\frac{x}{2}) c o s (\frac{x}{2})}$ $= \frac{- i s i n (\frac{(n + 1) x}{2}) e^{i \frac{(n + 1) x}{2}}}{- i s i n (\frac{x}{2}) e^{\frac{i x}{2}}} = \frac{s i n (\frac{(n + 1) x}{2}) e^{i \frac{n x}{2}}}{s i n (\frac{x}{2})}$ $= \frac{s i n (\frac{(n + 1) x}{2})}{s i n (\frac{x}{2})} {c o s (\frac{n x}{2}) + i s i n (\frac{n x}{2})} \Rightarrow$ $S_{n} (x) = \frac{s i n (\frac{(n + 1) x}{2}) c o s (\frac{n x}{2})}{s i n (\frac{x}{2})} a n d W_{n} (x) = \frac{s i n (\frac{(n + 1) x}{2}) s i n (\frac{n x}{2})}{s i n (\frac{x}{2})} \Rightarrow$ $1 + c o s x + c o s (2 x) + . . . c o s (13 x) = \frac{s i n (7 x) c o s (\frac{13 x}{2})}{s i n (\frac{x}{2})} a n d$ $1 + s i n x + s i n (2 x) + . . . + s i n (13 x) = 1 + \frac{s i n (7 x) s i n (\frac{13 x}{2})}{s i n (\frac{x}{2})} \Rightarrow$ $(. . .) \times (. . .) = \frac{s i n (7 x) c o s (\frac{13 x}{2})}{s i n (\frac{x}{2})} + \frac{{s i n}^{2} (7 x) s i n (13 x)}{2 {s i n}^{2} (\frac{x}{2})} \Rightarrow$ $\int_{- π}^{π} A (x) d x = \int_{- π}^{π} \frac{s i n (7 x) c o s (\frac{13 x}{2})}{s i n (\frac{x}{2})} d x + \int_{- π}^{π} \frac{{s i n}^{2} (7 x) s i n (13 x)}{2 {s i n}^{2} (\frac{x}{2})} d x$ $\int_{- π}^{π} \frac{s i n (7 x) c o s (\frac{13 x}{2})}{s i n (\frac{x}{2})} d x = 2 \int_{0}^{π} \frac{s i n (7 x) c o s (\frac{13 x}{2})}{s i n (\frac{x}{2})} d x (e v e n f u n c t i o n)$ $\int_{- π}^{π} \frac{{s i n}^{2} (7 x) s i n (13 x)}{2 {s i n}^{2} (\frac{x}{2})} d x = 0 (o d d f u n c t i o n) . . . b e c o n t i n u e d . . . .$
Commented by Meritguide1234 last updated on 13/Feb/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19

	$1) \int_{0}^{1} {(1 - x^{7})}^{\frac{1}{4}} d x - \int_{0}^{1} {(1 - x^{4})}^{\frac{1}{7}} d x$ $I_{1} - I_{2}$ $\int_{0}^{1} {(1 - x^{a})}^{\frac{1}{b}} d x$ $x^{a} = {s i n}^{2} α x = {(s i n α)}^{\frac{2}{a}}$ $d x = \frac{2}{a} \times {(s i n α)}^{\frac{2}{a} - 1} (c o s α) d α$ $\int_{0}^{\frac{π}{2}} {(1 - {s i n}^{2} α)}^{\frac{1}{b}} \times \frac{2}{a} \times {(s i n α)}^{\frac{2}{a} - 1} (c o s α) d α$ $\int_{0}^{\frac{π}{2}} {(c o s α)}^{\frac{2}{b}} \times \frac{2}{a} \times {(s i n α)}^{\frac{2}{a} - 1} (c o s α) d α$ $\frac{2}{a} \int_{0}^{\frac{π}{2}} {(s i n α)}^{\frac{2}{a} - 1} \times {(c o s α)}^{\frac{2}{b} + 1} d α$ $\frac{1}{a} \times 2 \int_{0}^{\frac{π}{2}} {(s i n α)}^{2 \times \frac{1}{a} - 1} {(c o s α)}^{2 (\frac{1}{b} + 1) - 1} d α$ $f o r m u l a$ $2 \int_{0}^{\frac{π}{2}} {(s i n α)}^{2 p - 1} {(c o s α)}^{2 q - 1} d α$ $\frac{⌈ (p) ⌈ q)}{⌈ (p + q)}$ $s o \frac{1}{a} \times 2 \int_{0}^{\frac{π}{2}} {(s i n α)}^{2 \times \frac{1}{a} - 1} {(c o s α)}^{2 (\frac{1}{b} + 1) - 1} d α$ $= \frac{1}{a} \times \frac{⌈ (\frac{1}{a}) ⌈ (\frac{1}{b} + 1)}{⌈ (\frac{1}{a} + \frac{1}{b} + 1)}$ $s o I_{1} = \frac{1}{7} \times \frac{⌈ (\frac{1}{7}) ⌈ (\frac{1}{4} + 1)}{⌈ (\frac{1}{4} + \frac{1}{7} + 1)} [a = 7 b = 4]$ $I_{2} = \frac{1}{4} \times \frac{⌈ (\frac{1}{4}) ⌈ (\frac{1}{7} + 1)}{⌈ (\frac{1}{4} + \frac{1}{7} + 1)} [a = 4 b = 7]$ $r e q u i r e d a n s = I_{1} - I_{2}$ $f o r m u l a ⌈ (n + 1) = n!$ $p l s c h e c k u p t o t h i s s t e p . . .$ $I_{1} - I_{2} = \frac{1}{⌈ (\frac{11}{28} + 1)} [\frac{1}{7} \times ⌈ (\frac{1}{7}) \times ⌈ (\frac{1}{4} + 1) - \frac{1}{4} \times ⌈ (\frac{1}{4}) \times ⌈ (\frac{1}{7} + 1)]$ $l e t a = \frac{1}{7} b = \frac{1}{4}$ $a n s = \frac{1}{⌈ (a + b + 1)} \times [a \times ⌈ a) ⌈ (b + 1) - b ⌈ (b) ⌈ (a + 1)]$ $= \frac{1}{(a + b)!} \times [a (a - 1)! b! - b (b - 1)! a!]$ $= \frac{1}{(a + b)!} [a! b! - b! a!] = 0$
	Commented by mr W last updated on 12/Feb/19

	$f (x) = {(1 - x^{7})}^{\frac{1}{4}} = y$ $\Rightarrow x = {(1 - y^{4})}^{\frac{1}{7}}$ $\Rightarrow f^{- 1} (x) = {(1 - x^{4})}^{\frac{1}{7}} = g (x)$ $\Rightarrow g (x) i s i n v e r s e f u n c t i o n o f f (x), i . e .$ $g (f (x)) = x$ $I_{2} = \int_{0}^{1} g (y) d y$ $l e t y = f (x)$ $\Rightarrow g (y) = g (f (x)) = x$ $d y = f^{'} (x) d x$ $\Rightarrow I_{2} = \int_{0}^{1} g (y) d y = \int_{g (0)}^{g (1)} {x f}^{'} (x) d x = \int_{1}^{0} {x f}^{'} (x) d x = - \int_{0}^{1} {x f}^{'} (x) d x$ $I_{1} = \int_{0}^{1} f (x) d x$ $I = I_{1} - I_{2} = \int_{0}^{1} f (x) d x + \int_{0}^{1} {x f}^{'} (x) d x$ $= \int_{0}^{1} [f (x) + {x f}^{'} (x)] d x$ $= {[x f (x)]}_{0}^{1}$ $= 0$
	Commented by rahul 19 last updated on 12/Feb/19

	$S i r, I^{'} m n e w t o t h i s f u n c t i o n!$
	Commented by rahul 19 last updated on 12/Feb/19

	Commented by rahul 19 last updated on 12/Feb/19

	$t h i s i s {s o l}^{n} g i v e n . P l e x p l a i n h o w$ $\int_{0}^{1} g (y) d y = \int_{1}^{0} x f^{'} (x) d x ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19

	$e x c e l l e n t s i r . . .$
	Commented by rahul 19 last updated on 13/Feb/19

	$T h a n k y o u s i r!$
	Answered by mr W last updated on 12/Feb/19

	$Q (2)$ $f (x) = x^{3} + 3 x + 4$ $I_{1} = \int_{- 1}^{1} f (x) d x = \int_{- 1}^{0} f (x) d x + \int_{0}^{1} f (x) d x$ $I_{2} = \int_{0}^{4} f^{- 1} (x) d x$ $l e t x = f (t) \Rightarrow t = f^{- 1} (x)$ $d x = f^{'} (t) d t$ $I_{2} = \int_{f^{- 1} (0)}^{f^{- 1} (4)} {t f}^{'} (t) d t = \int_{- 1}^{0} {t f}^{'} (t) d t = \int_{- 1}^{0} {x f}^{'} (x) d x$ $I = I_{1} + I_{2} = \int_{- 1}^{0} f (x) d x + \int_{0}^{1} f (x) d x + \int_{- 1}^{0} {x f}^{'} (x) d x$ $= \int_{- 1}^{0} [f (x) + {x f}^{'} (x)] d x + \int_{0}^{1} f (x) d x$ $= {[x f (x)]}_{- 1}^{0} + \int_{0}^{1} f (x) d x$ $= 0 + \int_{0}^{1} f (x) d x$ $= \int_{0}^{1} f (x) d x$ $= \int_{0}^{1} (x^{3} + 3 x + 4) d x$ $= {[\frac{x^{4}}{4} + \frac{3 x^{2}}{2} + 4 x]}_{0}^{1}$ $= \frac{1}{4} + \frac{3}{2} + 4$ $= \frac{23}{4}$ $\Rightarrow \int_{- 1}^{1} f (x) d x + \int_{0}^{4} f^{- 1} (x) d x = \frac{23}{4}$
	Commented by mr W last updated on 12/Feb/19

	$c e r t a i n l y o n e c a n a l s o d o l i k e t h i s :$ $I_{1} = \int_{- 1}^{1} f (x) d x = \int_{- 1}^{1} (x^{3} + 3 x + 4) d x$ $= {[\frac{x^{4}}{4} + \frac{3 x^{2}}{2} + 4 x]}_{- 1}^{1}$ $= 8$ $I_{2} = \int_{f^{- 1} (0)}^{f^{- 1} (4)} {t f}^{'} (t) d t = \int_{- 1}^{0} {t f}^{'} (t) d t = \int_{- 1}^{0} t (3 t^{2} + 3) d t$ $= 3 \int_{- 1}^{0} (t^{3} + t) d t$ $= 3 {[\frac{t^{4}}{4} + \frac{t^{2}}{2}]}_{- 1}^{0}$ $= - 3 (\frac{1}{4} + \frac{1}{2})$ $= - \frac{9}{4}$ $\Rightarrow I = 8 - \frac{9}{4} = \frac{23}{4}$
	Commented by rahul 19 last updated on 13/Feb/19

	$y e s, 2 n d m e t h o d w o r k s o n l y w h e n {f u n}^{c}$ $i s e a s y t o i n t e g r a t e!$
	Answered by mr W last updated on 12/Feb/19

	$Q (3)$ $I = \int_{- π}^{π} (1 + c o s x + c o s 2 x + . . . . + c o s 13 x) (1 + s i n x + . . . + s i n 13 x) d x$ $= \int_{- π}^{π} (1 + c o s x + c o s 2 x + . . . . + c o s 13 x) d x + \int_{- π}^{π} \underset{- - - - - - - - e v e n - - - - - - - - -}{(1 + c o s x + c o s 2 x + . . . . + c o s 13 x)} (\underset{- - - - - - o d d - - - - -}{s i n x + . . . + s i n 13 x}) d x$ $= \int_{- π}^{π} (1 + c o s x + c o s 2 x + . . . . + c o s 13 x) d x + 0$ $= 2 \int_{0}^{π} (1 + c o s x + c o s 2 x + . . . . + c o s 13 x) d x$ $= 2 {(x + \sin x + \frac{\sin 2 x}{2} + . . . + \frac{\sin 13 x}{13})}_{0}^{π}$ $= 2 π$
	Commented by rahul 19 last updated on 13/Feb/19
	thank you sir!
	Answered by mr W last updated on 13/Feb/19

	$Q (4)$ $l e t I_{2} = \int_{0}^{2} {(x^{2} + 2 x)}^{\frac{1}{3}} d x = \int_{0}^{2} {[{(x + 1)}^{2} - 1]}^{\frac{1}{3}} d (x + 1)$ $= \int_{1}^{3} {(t^{2} - 1)}^{\frac{1}{3}} d t$ $= \int_{1}^{3} {(x^{2} - 1)}^{\frac{1}{3}} d x = \int_{1}^{3} g (x) d x$ $w i t h g (x) = {(x^{2} - 1)}^{\frac{1}{3}}$ $l e t f (x) = \sqrt{x^{3} + 1} = y$ $x = {(y^{2} - 1)}^{\frac{1}{3}}$ $f^{- 1} (x) = {(x^{2} - 1)}^{\frac{1}{3}} = g (x)$ $l e t I_{1} = \int_{0}^{2} \sqrt{x^{3} + 1} d x = \int_{0}^{2} f (x) d x$ $l e t x = g (t)$ $f (x) = f (g (t)) = t$ $d x = g^{'} (t) d t$ $I_{1} = \int_{0}^{2} f (x) d x = \int_{f (0)}^{f (2)} {t g}^{'} (t) d t = \int_{1}^{3} {t g}^{'} (t) d t$ $= \int_{1}^{3} {x g}^{'} (x) d x$ $I = I_{1} + I_{2} = \int_{1}^{3} {x g}^{'} (x) d x + \int_{1}^{3} g (x) d x$ $= \int_{1}^{3} [g (x) + {x g}^{'} (x)] d x$ $= {[x g (x)]}_{1}^{3}$ $= 3 {(3^{2} - 1)}^{\frac{1}{3}} - 1 {(1^{2} - 1)}^{\frac{1}{3}}$ $= 3 \times 8^{\frac{1}{3}}$ $= 3 \times 2$ $= 6$ $\Rightarrow \int_{0}^{2} (\sqrt{x^{3} + 1} + {(x^{2} + 2 x)}^{\frac{1}{3}}) d x = 6$
	Commented by rahul 19 last updated on 14/Feb/19

	$T h a n k s s i r!$
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