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All Summer in a DayI would say that the resolution of the story is that there is no resolution! This might seem to not make sense, but the story ends on a kind of cliffhanger: Margot is locked in a closet, and misses...

The Most Dangerous GameGeneral Zaroff is Connell's antagonist in "The Most Dangerous Game." As villains normally do, Zaroff has a different perspective towards life and society than the rest of the world does. For...

The Outcasts of Poker FlatThe obvious answer to this question would be Mr. Oakhurst. He's the one who leads the group out of Poker Flat and into the mountains, hoping to speed them along before the weather hits. He's...

MathTo solve, let's apply the least common multiple (LCM) of a set of numbers. Pizza is served every sixth day. To get the days in which it will be served, take the multiples of 6. The multiples of 6...

The Hitchhiking GameFirst, it's important to determine the character of the young man before we begin our letter. Identifying the young man's temperament and personalitytype will help us to stay as true to character...

The Catcher in the RyeIt's interesting to view Jane's checker playing as a metaphor for her approach to life. Kings in the back row might stand for selfdiscipline, reserve, or a hidden ally. One thing that Holden...

MathGiven the equations y=0, x=6 and `y=\frac{x}{\sqrt{x+3}}` We have to find the area bounded by the three equations. The graph is shown below: The yellow region is the bounded area. Point (6,2) is...

MathGiven to solve , `int 4/(csc(theta)cot(theta)) d theta` just for the convenience let` x= theta` so,` int 4/(csc(theta)cot(theta)) d theta` =`int 4/(csc(x)cot(x)) d x` =`int...

Math`int (sin sqrt theta)/sqrt theta d theta` To solve, apply usubstitution method. `u=sqrt theta` `u= theta ^(1/2)` `du = 1/2 theta^(1/2) d theta` `du = 1/(2theta^(1/2))d theta` `du =1/(2 sqrt...

MathGiven to solve , `int cos(theta) / (1+cos(theta)) d theta` just for easy solving let `x=theta ` so the equation is given as `int cos(x) / (1+cos(x)) d x ` (1) let `u= tan(x/2)` ,=> then...

Math`int ( sin (theta))/(32cos(theta)) d theta` To solve, apply usubstitution method. `u = 32cos (theta)` `du = 2*(sin (theta)) d theta` `du = 2sin (theta) d theta` `1/2du= sin (theta) d theta`...

RikkiTikkiTaviAt first Nag introduces himself to RikkiTikki in a way designed to strike fear into the mongoose's heart and to make Rikki feel like a young thing. But Rikki's mongoose nature won't allow him to...

MathBasis (n=1) We will use integration by parts `int u dv=uvint v du` `int_0^infty xe^x dx=[u=x,dv=e^x dx],[du=dx,v=e^x]=` `xe^x_0^infty+int_0^infty e^x dx=(xe^xe^x)_0^infty=` `lim_(x...

Math`int_0^infty sin(x/2)dx=` Substitute `u=x/2` `=>` `du=dx/2` `=>` `dx=2du,` `u_l=0/2=0,` `u_l=infty/2=infty` (`u_l` and `u_u` are lower and upper bound respectively). `2int_0^infty sin u...

Math`int_0^infty cos (pi x)dx=` Substitute `u=pi x` `=>` `du=pi dx` `=>` `dx=(du)/pi,` `u_l=pi cdot 0=0,` `u_u=pi cdot infty=infty.` (`u_l` and `u_u` are lower and upper bound respectively)....

Math`int_0^infty e^x/(1+e^x)dx=` Substitute `u=1+e^x` `=>` `du=e^xdx,` `u_l=1+e^0=2,` `u_u=lim_(x to infty)(1+e^x)=infty.` `int_1^infty 1/u du=ln u_2^infty=lim_(u to infty)ln uln 2=infty` As...

Math`int_0^infty 1/(e^x+e^x)dx=` Multiply both numerator and the denominator by `e^x.` `int_0^infty e^x/(1+e^(2x))dx=` Substitute `u=e^x` `=>` `du=e^x dx,``u_l=e^0=1,``u_u=lim_(x to infty)e^x=infty...

Math`int_infty^infty 4/(16+x^2)dx=` Divide both numerator and denominator by 16. `int_infty^infty (1/4)/(1+x^2/16)dx=int_infty^infty (1/4 dx)/(1+(x/4)^2)=` Substitute `u=x/4` `=>` `du=1/4 dx.`...

Math`int_1^infty (ln x)/x dx=` Substitute `u=lnx` `=>` `du=1/x dx,` `u_l=ln 1=0,` `u_u=ln infty=infty` (`u_l` and `u_u` denote lower and upper bound respectively). `int_0^infty u...

Math`int_4^infty 1/(x(ln x)^3)dx=` Substitute `u=ln x` `=>` `du=1/x dx,` `u_l=ln 4,` `u_u=ln infty=infty` (`u_l` and `u_u` denote lower and upper bound respectively). `int_(ln 4)^infty1/u^3...

MathWe will use integration by parts (twice): `int u dv=uvint v du` `int_0^infty e^x cos x dx=[u=e^x,dv=cos x dx],[du=e^x dx,v=sin x]=` `e^x sin x+int_0^infty e^x sin x dx=[u=e^x,dv=sin x...

MathWe will use integration by parts `int u dv=uvint v du` We will need to apply integration by parts two times in order to eliminate `x^2` from under the integral. `int_0^infty x^2e^x...

MathWe will use integration by parts `int udv=uvint vdu` `int_0^infty xe^(x/3)dx=[u=x,dv=e^(x/3)dx],[du=dx,v=3e^(x/3)]=` `3xe^(x/3)_0^infty+3int_0^infty e^(x/3)dx=`...

MathWe will use integration by parts `int udv=uvint vdu` `int_infty^0 xe^(4x)dx=[u=x,dv=e^(4x)dx],[du=dx,v=1/4e^(4x)]=` `1/4xe^(4x)_infty^0+1/4int_infty^0 e^(4x)dx=`...

Math`int_1^infty 4/root(4)(x)dx=4int_1^infty x^(1/4)dx=4x^(3/4)/(3/4)_1^infty=16/3x^(3/4)_1^infty=` `16/3(lim_(x to infty) x^(3/4)1)=16/3(infty1)=infty` As we can see the value if the integral is...

Math`int_1^infty 3/root(3)(x)dx=` Rewrite the surd as a power using the following formula `root(n)(x^m)=x^(m/n).` `int_1^infty 3x^(1/3)dx=3x^(2/3)/(2/3)_1^infty=^(9x^(2/3))/2_1^infty=9/2( lim_(x to...

MathAn integral in which one of the limits of integration is infinity is an improper integral. Because we cannot find the definite integral using infinity (since it is not an actual value), we will...

MathThe function under the integral is continuous and bounded on any interval `[1, A].` The only point that makes the integral improper is `x=+oo.` Therefore this integral is the limit `lim_(A>+oo)...

MathAny integral with infinite bounds is an improper integral therefore this is an improper integral. `int_infty^0 e^(3x) dx=` Substitute `u=3x` `=>` `du=3dx,` `u_l=3cdot(infty)=infty,`...

MathThe integral is improper because the function under the integral is not defined at zero (see the image below). `1/sqrt0=1/0` `int_0^4 1/sqrt x dx=2sqrt x_0^4=2(sqrt 4sqrt0)=4` The integral...

MathIntegral `int_0^(pi/4)csc x dx` is improper because cosecant is not defined on zero (more generally `csc x` is not defined for `x in {k pi, k in ZZ}.`) and interval of integration includes that...

MathAny integral with infinite bounds is an improper integral. Hence `int_0^infty cos x dx` is an improper integral. Also, the cosine function is a periodic function and `lim_(x to infty)cos x` does...

MathThe integral does not have infinite bounds and the function is well defined over the whole interval of integration so there is no need to use limits. Therefore, the integral is not improper....

MathAny integral with infinite bounds is an improper integral, therefore `int_1^infty ln(x^2)dx` is an improper integral. Moreover, since the function under the integral is positive over the given...

MathIntegral `int_1^2 dx/x^3` is not an improper integral because function `f(x)=1/x^3` is well defined (with finite values) over the whole interval of integration `[1,2].` ` ` The only point where the...

MathGivne to solve, `lim_(x>1^(+)) (int_1^x cos(theta) d theta ) / (x1)` =`lim_(x>1^(+)) ([sin(theta)]_1^x) / (x1)` =`lim_(x>1^(+)) ([sin(x)sin(1)]) / (x1)` when `x> 1+` then...

Math`lim_(x>oo) (int_1^x ln(e^(4t1)) dt )/ x` = `lim_(x>oo) (int_1^x (4t1) dt )/ x` = `lim_(x>oo) ( [4(t^2)/2t]_1^x )/ x` = `lim_(x>oo) ( [2x^2 x][2(1^2)1] )/ x` = `...

MathGiven to solve, `lim_(x>0) x/arctan(2x)` as `x>0` then the `x/arctan(2x) =0/0` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general equation it is as...

MathGiven to solve, `lim_(x>0)arctanx/sinx` as `x>0 ` on substituting we get `arctanx/sinx = 0/0 ` so by using the L'hopital rule we get the solution as follows, as for the general equation it...

Math`lim_(x>1) (ln(x))/(sin(pix))` To solve, plugin x = 1. `lim_(x>1) (ln(x))/(sin(pix)) = (ln(1))/(sin(pi*1)) = 0/0` Since the result is indeterminate, to find the limit of the function as x...

Math`lim_(x>0) (sin (5x))/(tan(9x))` To solve, plugin x = 0. `lim_(x>0) (sin(5*0))/(tan(9*0)) = 0/0` Since the result is indeterminate, to find the limit of the function as x approaches zero,...

MathGiven to solve , `lim_(x>oo) e^(x/2)/x` As `x` thends to ` oo` we get `e^(x/2)/x = oo/oo` L'Hopital's Rule says if `lim_(x>a) f(x)/g(x) = 0/0` or `(+oo)/(+oo)` then the limit is:...

MathGiven to solve, `lim_(x>oo) e^x/(x^4)` as `x>oo` then the ` e^x/(x^4) =oo/oo` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general equation it is as...

MathGivne to solve , `lim_(x>oo)ln(x^4)/x^3` = `lim_(x>oo) 4 ln(x)/x^3` = `4lim_(x>oo) ln(x)/x^3` as `x>oo` then the `ln(x)/x^3 =oo/oo` form so upon applying the L 'Hopital rule we get...

MathGiven to solve, `lim_(x>oo) lnx/(x^2)` as `x>oo` then the `lnx/(x^2) =oo/oo` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general equation it is as...

MathGiven to solve, `lim_(x>oo) sinx/(xpi)` This can be solved by applying the squeeze theorem and is as follows as we know the limits or boundaries of `sin(x)` is `1<=sin(x)<=1` Dividing...

MathGiven to solve , `lim_(x>oo) cosx/x` by applying the squeeze theorem we can solve the limits and it is as follows, as we know the boundaries of the `cos(x)` as `1<=cos(x)<=1` now...

MathGiven `lim_(x>oo)x^2/sqrt(x^2+1)` as `x>oo` then we get `x^2/sqrt(x^2+1)=oo/oo` since it is of the form `oo/oo` , we can use the L 'Hopital rule so upon applying the L 'Hopital rule we...

MathGiven to solve `lim_(x>oo) (x/(sqrt(x+1)))` as `x> oo` we get `(x/(sqrt(x+1))) = oo/oo ` form so upon applying the L 'Hopital rule we get the solution as follows, as for the general...

MathGiven to solve , `lim_(x>oo) x^3/e^(x^2)` upon `x` tends to` oo` we get `x^3/e^(x^2) = oo/oo` so, by applying the L'Hopital Rule we get as for the general equation it is as follows...