f(x) = ((e^(3x) +e^(−3x) )/2) 1) determine f^(−1) (x) 2) calculate ∫_0 ^1 x f(x)dx and ∫_0 ^1 f(x)dx 3) calculate ∫ f^(−1) (x)dx 4) calculate u_n = ∫_0 ^π f(x)cos(nx)dx and v_n = ∫_0 ^n f(x)sin(nx)dx find nature of Σ (v_n /u_n ) ∫_0 ^1 xf(x) dx =(1/2) ∫_0 ^1 x e^(3x) dx +(1/2) ∫_0 ^1 x e^(−3x) dx (by parts) =(1/2){ [(x/3)e^(3x) ]_0 ^1 −(1/3)∫_0 ^1 e^(3x) dx +[−(x/3)e^(−3x) ]_0 ^1 +(1/3)∫_0 ^1 e^(−3x) dx} =(1/2){(e^3 /3) −(1/9)(e^3 −1) −(e^(−3) /3) −(1/9)(e^(−3) −1)}


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 42704 by maxmathsup by imad last updated on 01/Sep/18
$f(x) = ((e^(3x) +e^(−3x) )/2) 1) determine f^(−1) (x) 2) calculate ∫_0 ^1 x f(x)dx and ∫_0 ^1 f(x)dx 3) calculate ∫ f^(−1) (x)dx 4) calculate u_n = ∫_0 ^π f(x)cos(nx)dx and v_n = ∫_0 ^n f(x)sin(nx)dx find nature of Σ (v_n /u_n ) ∫_0 ^1 xf(x) dx =(1/2) ∫_0 ^1 x e^(3x) dx +(1/2) ∫_0 ^1 x e^(−3x) dx (by parts) =(1/2){ [(x/3)e^(3x) ]_0 ^1 −(1/3)∫_0 ^1 e^(3x) dx +[−(x/3)e^(−3x) ]_0 ^1 +(1/3)∫_0 ^1 e^(−3x) dx} =(1/2){(e^3 /3) −(1/9)(e^3 −1) −(e^(−3) /3) −(1/9)(e^(−3) −1)}$
$f (x) = \frac{e^{3 x} + e^{- 3 x}}{2}$ $1) d e t e r m i n e f^{- 1} (x)$ $2) c a l c u l a t e \int_{0}^{1} x f (x) d x a n d \int_{0}^{1} f (x) d x$ $3) c a l c u l a t e \int f^{- 1} (x) d x$ $4) c a l c u l a t e u_{n} = \int_{0}^{π} f (x) c o s (n x) d x a n d v_{n} = \int_{0}^{n} f (x) s i n (n x) d x$ $f i n d n a t u r e o f Σ \frac{v_{n}}{u_{n}}$ $\int_{0}^{1} x f (x) d x = \frac{1}{2} \int_{0}^{1} x e^{3 x} d x + \frac{1}{2} \int_{0}^{1} x e^{- 3 x} d x (b y p a r t s)$ $= \frac{1}{2} {{[\frac{x}{3} e^{3 x}]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} e^{3 x} d x + {[- \frac{x}{3} e^{- 3 x}]}_{0}^{1} + \frac{1}{3} \int_{0}^{1} e^{- 3 x} d x}$ $= \frac{1}{2} {\frac{e^{3}}{3} - \frac{1}{9} (e^{3} - 1) - \frac{e^{- 3}}{3} - \frac{1}{9} (e^{- 3} - 1)}$
Commented by maxmathsup by imad last updated on 01/Sep/18
$1) f(x)=ch(3x) =y ⇔ 3x =argch(y) ⇔x =(1/3)argch(y) =(1/3)ln(y+(√(y^2 −1))) ⇒ f^(−1) (x) =(1/3)ln(y+(√(y^2 −1))) 2)∫_0 ^1 f(x)dx = (1/2)∫_0 ^1 ( e^(3x) +e^(−3x) )dx =(1/2)[(1/3)e^(3x) −(1/3)e^(−3x) ]_0 ^1 =(1/6){e^3 −e^(−3) }$
$1) f (x) = c h (3 x) = y \Leftrightarrow 3 x = a r g c h (y) \Leftrightarrow x = \frac{1}{3} a r g c h (y) = \frac{1}{3} l n (y + \sqrt{y^{2} - 1}) \Rightarrow$ $f^{- 1} (x) = \frac{1}{3} l n (y + \sqrt{y^{2} - 1})$ $2) \int_{0}^{1} f (x) d x = \frac{1}{2} \int_{0}^{1} (e^{3 x} + e^{- 3 x}) d x = \frac{1}{2} {[\frac{1}{3} e^{3 x} - \frac{1}{3} e^{- 3 x}]}_{0}^{1}$ $= \frac{1}{6} {e^{3} - e^{- 3}}$
Commented by maxmathsup by imad last updated on 01/Sep/18

$3) l e t I = \int f^{- 1} (x) d x l e t f^{- 1} (x) = t \Rightarrow x = f (t) \Rightarrow$ $I = \int t f^{^{'}} (t) d t = t f (t) - \int f (t) d t a n d$ $\int f (t) d t = \int \frac{e^{3 t} + e^{- 3 t}}{2} d t = \frac{1}{6} e^{3 t} - \frac{1}{6} e^{- 3 t} + c \Rightarrow$ $I = t f (t) - \frac{1}{6} e^{3 t} + \frac{1}{6} e^{- 3 t} b u t t = f^{- 1} (x) = \frac{1}{3} l n (x + \sqrt{x^{2} - 1}) \Rightarrow$ $I = \frac{x}{3} l n (x + \sqrt{x^{2} - 1}) - \frac{1}{6} (x + \sqrt{x^{2} - 1}) + \frac{1}{6 (x + \sqrt{x^{2} - 1})} + c$
Commented by maxmathsup by imad last updated on 02/Sep/18
$4) we have u_n = ∫_0 ^π f(x)cos(nx)dx =(1/2) ∫_0 ^π (e^(3x) +e^(−3x) )cos(nx)dx ⇒ 2u_n = Re( ∫_0 ^π e^(inx) (e^(3x) +e^(−3x) )dx) =Re( ∫_0 ^π e^((3+in)x) + e^((−3+in)x) )dx but ∫_0 ^π ( e^((3+in)x) +e^((−3+in)x) )dx =[(1/(3+in)) e^((3+in)x) +(1/(−3+in)) e^((−3+in)x) ]_0 ^π =((e^(3π) (−1)^n )/(3+in)) +((e^(−3π) (−1)^n )/(−3 +in)) −(1/(3+in)) −(1/(−3+in)) =(((3−in) e^(3π) (−1)^n )/(9+n^2 )) + (((−3−in)e^(−3π) (−1)^n )/(9+n^2 )) −(1/(3+in)) +(1/(3−in)) =(((−1)^n )/(9+n^2 )) { 3 e^(3π) −in e^(3π) −3 e^(−3π) −in e^(−3π) } +(1/(3−in)) −(1/(3+in)) =((3(−1)^n )/(9+n^2 )){ e^(3π) −e^(−3π) } −((n(−1)^n )/(9+n^2 ))i {e^(3π) +e^(−3π) } +((2in)/(9+n^2 )) ⇒ u_n = ((3(−1)^n (e^(3π) −e^(−3π) ))/(9+n^2 )) also we have v_n = Im(∫_0 ^π e^((3+in)x) +e^((−3+in)x) )dx) ⇒v_n =((2n −n(−1)^n )/(9+n^2 )) .$
$4) w e h a v e u_{n} = \int_{0}^{π} f (x) c o s (n x) d x = \frac{1}{2} \int_{0}^{π} (e^{3 x} + e^{- 3 x}) c o s (n x) d x \Rightarrow$ $2 u_{n} = R e (\int_{0}^{π} e^{i n x} (e^{3 x} + e^{- 3 x}) d x) = R e (\int_{0}^{π} e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x b u t$ $\int_{0}^{π} (e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x = {[\frac{1}{3 + i n} e^{(3 + i n) x} + \frac{1}{- 3 + i n} e^{(- 3 + i n) x}]}_{0}^{π}$ $= \frac{e^{3 π} {(- 1)}^{n}}{3 + i n} + \frac{e^{- 3 π} {(- 1)}^{n}}{- 3 + i n} - \frac{1}{3 + i n} - \frac{1}{- 3 + i n}$ $= \frac{(3 - i n) e^{3 π} {(- 1)}^{n}}{9 + n^{2}} + \frac{(- 3 - i n) e^{- 3 π} {(- 1)}^{n}}{9 + n^{2}} - \frac{1}{3 + i n} + \frac{1}{3 - i n}$ $= \frac{{(- 1)}^{n}}{9 + n^{2}} {3 e^{3 π} - i n e^{3 π} - 3 e^{- 3 π} - i n e^{- 3 π}} + \frac{1}{3 - i n} - \frac{1}{3 + i n}$ $= \frac{3 {(- 1)}^{n}}{9 + n^{2}} {e^{3 π} - e^{- 3 π}} - \frac{n {(- 1)}^{n}}{9 + n^{2}} i {e^{3 π} + e^{- 3 π}} + \frac{2 i n}{9 + n^{2}}$ $\Rightarrow u_{n} = \frac{3 {(- 1)}^{n} (e^{3 π} - e^{- 3 π})}{9 + n^{2}} a l s o w e h a v e v_{n} = I m (\int_{0}^{π} e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x)$ $\Rightarrow v_{n} = \frac{2 n - n {(- 1)}^{n}}{9 + n^{2}} .$
Commented by maxmathsup by imad last updated on 02/Sep/18
$u_n =((3(−1)^n (e^(3π) −e^(−3π) ))/(2(9+n^2 ))) and v_n = ((2n−n(−1)^n { e^(3π) +e^(−3π) })/(2(9+n^2 )))$
$u_{n} = \frac{3 {(- 1)}^{n} (e^{3 π} - e^{- 3 π})}{2 (9 + n^{2})} a n d v_{n} = \frac{2 n - n {(- 1)}^{n} {e^{3 π} + e^{- 3 π}}}{2 (9 + n^{2})}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum