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IntegrationQuestion and Answers: Page 241

Question Number 55638 Answers: 1 Comments: 0

For all n ∈ N f_n (x)= { ((((nx)/(2n−1)), x ∈ [0, ((2n−1)/n)])),((1 , x ∈[((2n−1)/n), 2])) :} then for n→∞ ∫_1 ^2 f_n (x) dx convergences to..

$For all n \in N$ $f_{n} (x) = {\begin{cases} \frac{n x}{2 n - 1}, x \in [0, \frac{2 n - 1}{n}] \\ 1, x \in [\frac{2 n - 1}{n}, 2] \end{cases}$ $then for n \to \infty$ $\int_{1}^{2} f_{n} (x) d x convergences to . .$

Question Number 55615 Answers: 1 Comments: 0

let F(α)=∫_α ^(1+α^2 ) ((sin(αx))/(1+αx^2 ))dx 1) calculate (dF/dα)(α) 2) calculate lim_(α→0) F(α)

$l e t F (α) = \int_{α}^{1 + α^{2}} \frac{s i n (α x)}{1 + α x^{2}} d x$ $1) c a l c u l a t e \frac{d F}{d α} (α)$ $2) c a l c u l a t e {l i m}_{α \to 0} F (α)$

Question Number 55571 Answers: 0 Comments: 1

let u_n = ∫_(π/(n+1)) ^(π/n) (√(tan(x)))dx with n≥3 1) calculate U_n interms of n and calculate lim_(n→+∞ ) U_n 2) find nature of the serie Σ_(n≥3) U_n

$l e t u_{n} = \int_{\frac{π}{n + 1}}^{\frac{π}{n}} \sqrt{t a n (x)} d x w i t h n ⩾ 3$ $1) c a l c u l a t e U_{n} i n t e r m s o f n a n d c a l c u l a t e {l i m}_{n \to + \infty} U_{n}$ $2) f i n d n a t u r e o f t h e s e r i e \sum_{n ⩾ 3} U_{n}$

Question Number 55560 Answers: 1 Comments: 0

Question Number 55577 Answers: 1 Comments: 1

Question Number 55520 Answers: 3 Comments: 2

How can solve ∫(√)tan(x)dx ?

$H o w c a n s o l v e \int \sqrt{} \tan (x) d x ?$

Question Number 55467 Answers: 0 Comments: 3

Question Number 55457 Answers: 0 Comments: 2

calculate ∫_0 ^∞ ((ln(x))/(x^2 +x+1))dx .

$c a l c u l a t e \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x .$

Question Number 55364 Answers: 0 Comments: 0

prove that ∫_(−∞ ) ^∞ f(x)dx=1 such that f(x)=(1/((√n) β((n/2),(1/2))))(1+(x^2 /n))^(−(1/2)(1+n)) and β((n/2),(1/2))=∫_0 ^∞ (x^((n/2)−1) /((1+x)^(3/2) ))dx

$prove that \int_{- \infty}^{\infty} f (x) dx = 1$ $such that f (x) = \frac{1}{\sqrt{n} β (\frac{n}{2}, \frac{1}{2})} {(1 + \frac{x^{2}}{n})}^{- \frac{1}{2} (1 + n)}$ $and β (\frac{n}{2}, \frac{1}{2}) = \int_{0}^{\infty} \frac{x^{\frac{n}{2} - 1}}{{(1 + x)}^{\frac{3}{2}}} dx$

Question Number 55360 Answers: 2 Comments: 1

∫_1 ^( 2) (√(sin (3x−x^2 −2)))dx + (1/2)∫_3 ^1 (√(sin(((4t−t^2 −3)/4))))dt =?

$\int_{1}^{2} \sqrt{\sin (3 x - x^{2} - 2)} d x + \frac{1}{2} \int_{3}^{1} \sqrt{s i n (\frac{4 t - t^{2} - 3}{4})} d t = ?$

Question Number 55310 Answers: 1 Comments: 1

∫ 3x(√( 3x^3 + 7)) dx = . . . .

$\int 3 x \sqrt{3 x^{3} + 7} d x = . . . .$

Question Number 55282 Answers: 1 Comments: 1

calculatef(a)= ∫ (1+(a/x^2 ))arctan((a/x))dx 2) calculate ∫_1 ^(+∞) (1+(2/x^2 ))arctan((2/x))dx .

$c a l c u l a t e f (a) = \int (1 + \frac{a}{x^{2}}) a r c t a n (\frac{a}{x}) d x$ $2) c a l c u l a t e \int_{1}^{+ \infty} (1 + \frac{2}{x^{2}}) a r c t a n (\frac{2}{x}) d x .$

Question Number 55280 Answers: 1 Comments: 1

fint f(t)=∫_0 ^1 ((ln(1+tx^2 ))/x^2 )dx .

$f i n t f (t) = \int_{0}^{1} \frac{l n (1 + {t x}^{2})}{x^{2}} d x .$

Question Number 55279 Answers: 0 Comments: 0

find L( e^(−x) ln(1+x^2 )) with L mean laplace transform

$f i n d L (e^{- x} l n (1 + x^{2})) w i t h L m e a n l a p l a c e$ $t r a n s f o r m$

Question Number 55278 Answers: 0 Comments: 0

calculate ∫_0 ^(+∞) (dx/((x+1)(x+2)....(x+n)))

$c a l c u l a t e \int_{0}^{+ \infty} \frac{d x}{(x + 1) (x + 2) . . . . (x + n)}$

Question Number 55274 Answers: 0 Comments: 4

let ϕ(a) =∫_1 ^(√3) arctan((a/x))dx 1) calculate ϕ(a) interms of a 2) calculate ϕ^′ (a) at form of integral. 3) determine ϕ^((n)) (a) at form of integral. 4) find the value of ∫_1 ^(√3) arctan((2/x))dx .

$l e t φ (a) = \int_{1}^{\sqrt{3}} a r c t a n (\frac{a}{x}) d x$ $1) c a l c u l a t e φ (a) i n t e r m s o f a$ $2) c a l c u l a t e φ^{^{'}} (a) a t f o r m o f i n t e g r a l .$ $3) d e t e r m i n e φ^{(n)} (a) a t f o r m o f i n t e g r a l .$ $4) f i n d t h e v a l u e o f \int_{1}^{\sqrt{3}} a r c t a n (\frac{2}{x}) d x .$

Question Number 55273 Answers: 0 Comments: 1

let f(x) =∫_x^2 ^(1+x) (dt/(1+t+t^2 )) 1) calculate f(x) interms of x 2) calculate lim_(x→0) f(x) and lim_(x→+∞) f(x)

$l e t f (x) = \int_{x^{2}}^{1 + x} \frac{d t}{1 + t + t^{2}}$ $1) c a l c u l a t e f (x) i n t e r m s o f x$ $2) c a l c u l a t e {l i m}_{x \to 0} f (x) a n d {l i m}_{x \to + \infty} f (x)$

Question Number 55271 Answers: 0 Comments: 0

1) let f(x) =∫_0 ^(2π) ((cost)/(3 +sin(xt)))dt find a explicit form of f(x) 2) calculate g(x) =∫_0 ^(2π) ((tcos(xt)cost)/((3 +sin(xt))^2 ))dt 3) calculate ∫_0 ^(2π) ((cost)/(3+sint)) and ∫_0 ^(2π) ((t cos^2 t)/((3+sint)^2 ))dt

$1) l e t f (x) = \int_{0}^{2 π} \frac{c o s t}{3 + s i n (x t)} d t$ $f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e g (x) = \int_{0}^{2 π} \frac{t c o s (x t) c o s t}{{(3 + s i n (x t))}^{2}} d t$ $3) c a l c u l a t e \int_{0}^{2 π} \frac{c o s t}{3 + s i n t} a n d \int_{0}^{2 π} \frac{t {c o s}^{2} t}{{(3 + s i n t)}^{2}} d t$

Question Number 55267 Answers: 0 Comments: 0

1) calculate f(x)=∫_0 ^(π/4) ln(1+xtanθ)dθ 2) find the values of integrals ∫_0 ^(π/4) ln(1+tanθ) and ∫_0 ^(π/4) ln(1+2tanθ)dθ . 1) we have f^′ (x)=∫_0 ^(π/4) ((tanθ)/(1+xtanθ)) dθ =∫_0 ^(π/4) (((sinθ)/(cosθ))/(1+x((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) ((sinθ)/(cosθ +xsinθ)) dθ =_(tan((θ/2))=t) ∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 )) +((2xt)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((1+t^2 )(1−t^2 +2xt)))dt =−∫_0 ^((√2)−1) ((4t)/((t^2 +1)(t^2 −2xt −1)))dt let decompose F(t) = ((4t)/((t^2 +1)(t^2 −2xt −1))) roots of t^2 −2xt −1 Δ^′ =x^2 +1 ⇒t_1 =x+(√(x^2 +1)) and t_2 =x−(√(x^2 +1)) F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t)=((4t_1 )/((t_1 ^2 +1)(t_1 −t_2 ))) =α b =lim_(t→t_2 ) (t−t_2 )F(t) =((4t_2 )/((t_2 ^2 +1)(t_2 −t_1 ))) =β ⇒F(t)=(α/(t−t_1 )) +(β/(t−t_2 )) +((ct +d)/(t^2 +1)) F(0) =0=−(α/t_1 ) −(β/t_2 ) +d ⇒d =(α/t_1 ) +(β/t_2 ) F(1)=(2/(−2x)) =−(1/x)=(α/(1−t_1 )) +(β/(1−t_2 )) +((c+d)/2) ⇒(1/x) =(α/(t_1 −1)) +(β/(t_2 −1)) −(c/2) −(d/2) ⇒(c/2) =(α/(t_1 −1)) +(β/(t_2 −1)) −(d/2) −(1/x) ⇒c =((2α)/(t_1 −1)) +((2β)/(t_2 −1)) −d−(2/x) ∫ F(t)dt =αln∣t−t_1 ∣ +βln∣t−t_2 ∣ +(c/2)ln(t^2 +1) +d arctan(t) ⇒ ∫_0 ^((√2)−1) F(t)dt =[αln∣t−t_1 ∣+βln∣t−t_2 ∣ +(c/2)ln(t^2 +1)]_0 ^((√2)−1) =αln∣(√2)−1−t_1 ∣ +βln∣(√2)−1−t_2 ∣ +(c/2)ln(4−2(√2)) =αln∣(√2)−1−x−(√(1+x^2 )))+βln∣(√2)−1−x+(√(1+x^2 ))) +((ln(4−2(√2)))/2)c =f^′ (x) ⇒ f(x)=∫ αln∣(√2)−1−x−(√(1+x^2 ))∣)dx+β∫ ln∣(√2)−1+(√(1+x^2 ))∣dx +((cx)/2)ln(4−2(√2)) +C ....be continued...

$1) c a l c u l a t e f (x) = \int_{0}^{\frac{π}{4}} l n (1 + x t a n θ) d θ$ $2) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\frac{π}{4}} l n (1 + t a n θ) a n d \int_{0}^{\frac{π}{4}} l n (1 + 2 t a n θ) d θ .$ $1) w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{t a n θ}{1 + x t a n θ} d θ = \int_{0}^{\frac{π}{4}} \frac{\frac{s i n θ}{c o s θ}}{1 + x \frac{s i n θ}{c o s θ}} d θ$ $= \int_{0}^{\frac{π}{4}} \frac{s i n θ}{c o s θ + x s i n θ} d θ =_{t a n (\frac{θ}{2}) = t} \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 x t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) (1 - t^{2} + 2 x t)} d t = - \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(t^{2} + 1) (t^{2} - 2 x t - 1)} d t l e t d e c o m p o s e$ $F (t) = \frac{4 t}{(t^{2} + 1) (t^{2} - 2 x t - 1)} r o o t s o f t^{2} - 2 x t - 1$ $Δ^{^{'}} = x^{2} + 1 \Rightarrow t_{1} = x + \sqrt{x^{2} + 1} a n d t_{2} = x - \sqrt{x^{2} + 1}$ $F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{4 t_{1}}{(t_{1}^{2} + 1) (t_{1} - t_{2})} = α$ $b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{4 t_{2}}{(t_{2}^{2} + 1) (t_{2} - t_{1})} = β \Rightarrow F (t) = \frac{α}{t - t_{1}} + \frac{β}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $F (0) = 0 = - \frac{α}{t_{1}} - \frac{β}{t_{2}} + d \Rightarrow d = \frac{α}{t_{1}} + \frac{β}{t_{2}}$ $F (1) = \frac{2}{- 2 x} = - \frac{1}{x} = \frac{α}{1 - t_{1}} + \frac{β}{1 - t_{2}} + \frac{c + d}{2} \Rightarrow \frac{1}{x} = \frac{α}{t_{1} - 1} + \frac{β}{t_{2} - 1} - \frac{c}{2} - \frac{d}{2}$ $\Rightarrow \frac{c}{2} = \frac{α}{t_{1} - 1} + \frac{β}{t_{2} - 1} - \frac{d}{2} - \frac{1}{x} \Rightarrow c = \frac{2 α}{t_{1} - 1} + \frac{2 β}{t_{2} - 1} - d - \frac{2}{x}$ $\int F (t) d t = α l n ∣ t - t_{1} ∣ + β l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t) \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} F (t) d t = {[α l n ∣ t - t_{1} ∣ + β l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1)]}_{0}^{\sqrt{2} - 1}$ $= α l n ∣ \sqrt{2} - 1 - t_{1} ∣ + β l n ∣ \sqrt{2} - 1 - t_{2} ∣ + \frac{c}{2} l n (4 - 2 \sqrt{2})$ $= α l n ∣ \sqrt{2} - 1 - x - \sqrt{1 + x^{2}}) + β l n ∣ \sqrt{2} - 1 - x + \sqrt{1 + x^{2}}) + \frac{l n (4 - 2 \sqrt{2})}{2} c = f^{^{'}} (x) \Rightarrow$ $f (x) = \int α l n ∣ \sqrt{2} - 1 - x - \sqrt{1 + x^{2}} ∣) d x + β \int l n ∣ \sqrt{2} - 1 + \sqrt{1 + x^{2}} ∣ d x$ $+ \frac{c x}{2} l n (4 - 2 \sqrt{2}) + C . . . . b e c o n t i n u e d . . .$