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3

Answer:

We need to split up the integral.

Explanation:

Recall that #absu = {(u,"if",u >= 0),(-u,"if",u<0):}#

so #abs(sinx) = {(sinx,"if",sinx >= 0),(-sinx,"if",sinx < 0):}#.

We are integrating on #[0,(3pi)/2]# and we know that

#{(sinx >=0,"if" ,0 <=x <= pi),(sinx < 0,"if",pi < x <= (3pi)/2):}#

Therefore,

#abs(sinx) = {(sinx,"if",0 <= x <= pi),(-sinx,"if",pi < x < (3pi)/2):}#.

#int_0^((3pi)/2) 5abs(sinx) dx = 5int_0^((3pi)/2) abs(sinx) dx #

# = 5[int_0^pi sinx dx + int_pi^((3pi)/2) -sinx dx]#

# = 5[int_0^pi sinx dx - int_pi^((3pi)/2) sinx dx]#

Now use the fact that #int sinx dx = -cosx +C# to find each of the integrals.

# = 5[{:-cosx]_0^pi +{:cosx]_pi^((3pi)/2)]#

# = 5[(-cospi+cos0)+(cos((3pi)/2)-cospi)]#

# = 5[(-(-1)+1+0-(-1)]#

# = 5[3] = 15#

Bonus method

Some people prefer to integrate #abs(f(x))# by simply integrating from one zero to the next without first adjusting the sign. Any integral that comes out negative, we make positive.

The notation for this technique is

#int_0^((3pi)/2) 5abs(sinx) dx = abs(int_0^pi 5sinx dx)+abs(int_pi^((3pi)/2) 5sinx dx)#

The first of these two integrals will be positive and the second will be negative. (That's why the first method changed the sign for the second integral before integrating.)

2

Answer:

Because the order of the numerator is greater than the denominator some reduction must be done before expanding the partial fractions. Please see the explanation.

Explanation:

Given: #(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2)#

Begin reducing by adding 0 to the numerator in the form of #-2x^2 + 2x^2#

#(x^4 + x^3 -2x^2 + 2x^2 + x^2 + 1)/(x^2 + x - 2)#

Break into two fractions:

#(x^4 + x^3 -2x^2)/(x^2 + x - 2) + (2x^2 + x^2 + 1)/(x^2 + x - 2)#

Factor #x^2# from the numerator of the first term:

#(x^2(cancel(x^2 + x -2)))/cancel(x^2 + x - 2) + (2x^2 + x^2 + 1)/(x^2 + x - 2)#

The first term becomes #x^2# and we combine like terms in the numerator of the second fraction:

#x^2 + (3x^2 + 1)/(x^2 + x - 2)#

Add 0 to the numerator of the second term in the form of #3x - 6 - 3x + 6#:

#x^2 + (3x^2 + 3x - 6 - 3x + 6 + 1)/(x^2 + x - 2)#

Break the second term into two fractions:

#x^2 + (3x^2 + 3x - 6)/(x^2 + x - 2) + (-3x + 6 + 1)/(x^2 + x - 2)#

Remove a factor of 3 from the first numerator and combine like terms in the last fraction:

#x^2 + (3(cancel(x^2 + x - 2)))/cancel(x^2 + x - 2) + (7 -3x)/(x^2 + x - 2)#

The second term becomes 3:

#x^2 + 3 + (7 -3x)/(x^2 + x - 2)#

Partial Fraction Expansion of the last term:

#(7 -3x)/(x^2 + x - 2) = A/(x + 2) + B/(x - 1)#

#7 - 3x = A(x - 1) + B(x + 2)#

Make B disappear by letting x = -2

#7 - 3(-2) = A(-2 - 1) + B(-2 + 2)#

#A# = -13/3

Make A disappear by Letting x = 1:

#7 - 3(1) = A(1 - 1) + B(1 + 2)#

#B = 4/3#

Check

#(-13/3)/(x + 2) + (4/3)/(x - 1)#

#(-13/3)/(x + 2)(x - 1)/(x - 1) + (4/3)/(x - 1)(x + 2)/(x + 2)#

#((-13/3)(x - 1)+ (4/3)(x + 2))/((x + 2)(x - 1))#

#(7 - 3x)/((x + 2)(x - 1))#

This checks

Returning to the main problem:

#int(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2)dx = intx^2dx + 3intdx - 13/3int1/(x + 2)dx + 4/3int1/(x - 1)dx#

#int(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2)dx = 1/3x^3 + 3x - 13/3ln(x + 2) + 4/3ln(x - 1) + C#

1

Answer:

The integral is #ln|(1 + sqrt(1 - x^2))/x| - cos^-1x/x + C#

Explanation:

This can be rewritten as #intcos^-1x x^-2dx#. Whenever you see a product within an integral, you should consider using integration by parts.

The integration by parts formula states that an integral #intudv = uv - int(v du)#. #cos^-1x# is not that simple to integrate, so let's say #u = cos^-1x# and #dv= x^-2#. Then #du = -1/sqrt(1 - x^2)dx# and #v = -1/x#.

Use the formula now:

#intcos^-1x x^-2dx = -cos^-1x/x - int(-1/x * -1/sqrt(1 - x^2)dx)#

#intcos^-1x x^-2dx = -cos^-1x/x - int 1/(xsqrt(1 - x^2)) dx#

The remaining integral will have to be integrated by trigonometric substitution. This is a technique of integration where you make use of a pythagorean identity to get rid of the #√#. When the square root within the integral is of the form #sqrt(a^2 - x^2)# like the one above, use the substitution #color(red)(x = sin theta)#. Then #dx = costheta d theta#.

#int1/(xsqrt(1 - x^2)) dx= int1/(sinthetasqrt(1 - sin^2theta)) * costheta d theta#

#int1/(xsqrt(1 - x^2))dx =int 1/(sin thetasqrt(cos^2theta)) *costheta d theta#

#int1/(xsqrt(1 - x^2))dx = int1/sintheta d theta#

#int 1/(xsqrt(1 - x^2))dx = intcsctheta d theta#

This is a known integral that is derived here.

#int 1/(xsqrt(1 - x^2))dx = -ln|csctheta + cottheta| + C#

We now reverse the substitutions by drawing a triangle and finding the correct ratios. We know from our original substitution that #x/1 = sintheta#.

enter image source here

#int 1/(xsqrt(1 - x^2))dx = -ln|1/x + sqrt(1 - x^2)/x| + C#

#int 1/(xsqrt(1 - x^2))dx = -ln|(1 + sqrt(1 - x^2))/x| + C#

Let's return our attention to the whole integral.

#intcos^-1x x^-2dx = -cos^-1x/x- (- ln|(1 + sqrt(1 - x^2))/x|) + C#

#intcos^-1x x^-2dx = ln|(1 + sqrt(1 - x^2))/x| - cos^-1x/x + C#

Hopefully this helps!

1

Answer:

#1/2#

Explanation:

Before evaluating, we always have to find the integral. We look to simplify the integral as much as possible to see if anything can be cancelled.

#int_0^1 (x(x + 2))/(x + 1)^2#

We can't cancel anything, but a u-substitution would be effective. Let #u = x + 1#. Then #du = dx#. Also note that #x = u - 1# and #x + 2 = u + 1#. The integral therefore becomes:

#int_0^1 ((u - 1)(u + 1))/u^2 du#

Expand this:

#int_0^1 (u^2 - u + u - 1)/u^2 du#

#int_0^1 (u^2 - 1)/u^2 du#

Break into separate fractions.

#int_0^1 u^2/u^2 - 1/u^2 du#

#int_0^1 1 - 1/u^2 du#

#int_0^1 1 - u^-2 du#

You can now integrate this as #intx^ndx = x^(n + 1)/(n + 1) + C#, where #n != -1#.

#[u + 1/u]_0^1#

Reverse the substitution, since the initial variable wasn't #u#, it was #x#.

#[x + 1 + 1/(x + 1)]_0^1#

Evaluate using the second fundamental theorem of calculus, which states that #int_a^b F(x)dx = f(b) - f(a)#, where #f'(x) = F(x)# in all #[a, b]#.

#1 + 1 + 1/(1 + 1) - (0 + 1 + 1/(0 + 1))#

#2 + 1/2 - 2#

#1/2#

Hopefully this helps!

1

Answer:

#-lnx/(1+x)+lnx-ln(1+x)+C#

Explanation:

#I=intlnx/(1+x)^2dx#

Integration by parts takes the form #intudv=uv-intvdu#. Let:

#u=lnx#
#dv=1/(1+x)^2dx#

Differentiate #u# and integrate #dv#. The integration of #dv# is best performed with the substitution #t=1+x=>dt=dx#. You should get:

#du=1/xdx#
#v=-1/(1+x)#

Then:

#I=uv-intvdu#

#I=-lnx/(1+x)-int1/x(-1/(1+x))dx#

#I=-lnx/(1+x)+int1/(x(1+x))dx#

Perform partial fraction decomposition on #1/(x(1+x))#:

#1/(x(1+x))=A/x+B/(1+x)#

Then:

#1=A(1+x)+Bx#

Letting #x=-1#:

#1=A(1-1)+B(-1)#

#B=-1#

Letting #x=0#:

#1=A(1+0)+B(0)#

#1=A#

Then:

#1/(x(1+x))=1/x-1/(1+x)#

So:

#I=-lnx/(1+x)+int1/xdx-int1/(1+x)dx#

These are simple integrals. The second can be performed with the substitution #s=1+x=>ds=dx#.

#I=-lnx/(1+x)+lnx-ln(1+x)+C#

3

I got:

#mu(y) = e^y#


The point of an integrating factor is to turn an inexact differential into an exact one. One physical application of this is to turn a path function into a state function in chemistry (such as dividing by #T# to turn #q_"rev"#, a path function, into #S#, a state function, entropy).

I assume that the second terms include #3y^2#, not #3y# (it would be odd to not simply write #9y#).

Two options to find the special integrating factor, as defined by Nagle, are:

#bb(mu(x) = "exp"[int (((delM)/(dely))_x - ((delN)/(delx))_y)/(N(x,y))dx])#,

#bb(mu(y) = "exp"[int (((delN)/(delx))_y - ((delM)/(dely))_x)/(M(x,y))dy])#,

for the differential

#bb(dF(x,y) = ((delF)/(delx))_y dx + ((delF)/(dely))_xdy)#,

where #M = ((delF)/(delx))_y# and #N = ((delF)/(dely))_x#.

For now, let's find the partial derivatives. For your differential:

#color(green)(M(x,y) = 3x^2 + 3y^2)#
#color(green)(N(x,y) = x^3 + 3xy^2 + 6xy)#

Therefore:

#color(green)(((delM)/(dely))_x = 6y)#

#color(green)(((delN)/(delx))_y = 3x^2 + 3y^2 + 6y)#

which are clearly not equal, so the current differential is inexact. So, let us divide by #N(x,y)# and see if the integral with respect to #x# is reasonable to do:

#lnmu(x) = int (6y - 3x^2 - 3y^2 - 6y)/(x^3 + 3xy^2 + 6xy)dx#

#= int (-3x^2 - 3y^2)/(x^3 + 3xy^2 + 6xy)dx#

This doesn't look all that nice (it cannot be readily factored to eliminate #y# terms), so what if we try integrating with respect to #y# instead, and dividing by #M(x,y)# instead? Then:

#mu(y) = "exp"[int (((delN)/(delx))_y - ((delM)/(dely))_x)/(M(x,y))dy]#

and:

#lnmu(y) = int(3x^2 + 3y^2 + 6y - 6y)/(3x^2 + 3y^2)dy#

#= intcancel((3x^2 + 3y^2)/(3x^2 + 3y^2))^(1)dy#

And this integral is easy. It's just #y#. Therefore, #color(blue)(mu(y) = e^y)# would be your integrating factor. Let's test it out:

#3e^y(x^2 + y^2)dx + xe^y(x^2 + 3y^2 + 6y)dy = 0#

#(3x^2e^y + 3y^2e^y)dx + (x^3e^y + 3xy^2e^y + 6xye^y)dy = 0#

Checking for exactness, we obtain:

#((delM)/(dely))_x stackrel(?" ")(=) ((delN)/(delx))_y#

#3x^2e^y + 3(y^2e^y + 2ye^y) stackrel(?" ")(=) 3x^2e^y + 3y^2e^y + 6ye^y#

#3x^2e^y + 3y^2e^y + 6ye^y = 3x^2e^y + 3y^2e^y + 6ye^y# #color(blue)(sqrt"")#

so we know our integrating factor is correct!

2

Answer:

# x-y-8z =-16 #

Explanation:

First we rearrange the equation of the surface into the form # f(x,y,z)=0#

# x=y(2z-3) #
# :. x=2yz-3y #
# :. x+3y-2yz = 0 #

And so we have our function:

# f(x,y,z) = x+3y-2yz #

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

# grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k) #

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

# grad f = ((partial)/(partial x) (x+3y-2yz))hat(i) + #
# " " ((partial)/(partial y) (x+3y-2yz))hat(j) + #
# " " ((partial)/(partial z) (x+3y-2yz))hat(k) #
# " "= (1)hat(i) + (3-2z)hat(j) + (-2y)hat(k) #
# " "= hat(i) + (3-2z)hat(j) -(2y)hat(k) #

So for the particular point #(4,4,2)# the normal vector to the surface is given by:

# grad f(4,4,2) = hat(i) + (3-2*2)hat(j) -(2*4)hat(k) #
# " " = hat(i) -hat(j) -8hat(k) #
# " " = ( (1), (-1),(-8) ) #

So the tangent plane to the surface # x=y(2z-3) # has this normal vector and it also passes though the point #(4,4,2)#. It will therefore have a vector equation of the form:

# vec r * vec n = vec a * vec n #

Where #vec r=((x),(y),(z))#; #vec n# is the normal, and #a# is any point in the plane

Hence, the tangent plane equation is:

# ((x),(y),(z)) * ( (1), (-1),(-8) ) = ((4),(4),(2)) * ( (1), (-1),(-8) ) #
# :. (x)(1) + (y)(-1) + (z)(-8) = (4)(1) + (4)(-1) + (2)(-8) #
# :. x-y-8z = 4-4-16 #
# :. x-y-8z =-16 #

We can confirm this graphically: Here is the surface with the normal vector:
enter image source here

and here is the surface with the tangent plane and the normal vector:
enter image source here

1

Answer:

# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

Where #erf(x)# is the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Explanation:

# dy/dx = 1 + 2xy #
# :. dy/dx - 2xy = 1 # ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

# I = e^(int P(x) dx)#
# \ \ = e^(int \ -2x \ dx)#
# \ \ = e^(-x^2) #

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

# dy/dx - 2xy = 1 #
# :. e^(-x^2)dy/dx - 2xye^(-x^2) = 1*e^(-x^2) #
# :. d/dx(ye^(-x^2)) = e^(-x^2) #

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

# ye^(-x^2) = int \ e^(-x^2) \ dx#

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Which gives us:

# ye^(-x^2) = sqrt(pi)/2erf(x) + A#
# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

2

Answer:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

Explanation:

enter image source here

Our parabola #y=x^2# is parametrised by #x=t# and #y=t^2#, and #A(t,t^2)# is a general point on the parabola, and we give the point #P# the coordinates (to be determined) #P(alpha, beta)#

Differentiating wrt #x# we have:

#dy/dx=2x = 2t #

So the gradient of the tangent at #A# is given by #dy/dx=t#, We are told that #AP# is perpendicular to to parabola, in other words, it is perpendicular to the tangent at A. Hence the gradient of AP is #-1/(2t)# as the product of their gradients are -1.

We can now form the equation of the normal #AP# using the point-slope form of the straight line, #y-y_1=m(x-x_1)#:

# y-t^2=-1/(2t)(x-t) #
# :. 2ty-2t^3=-(x-t) #
# :. 2ty-2t^3=+t-x #

#P(alpha, beta)# lies on this line, giving:

# :. 2tbeta-2t^3 = t -alpha#

And we are also told that #AP# has length 1, so by Pythagoras:

# (t-alpha)^2+(t^2-beta)^2=1^2 #

We can combine these equations to eliminate #t -alpha#:

# (2tbeta-2t^3)^2+(t^2-beta)^2=1 #
# :. 4t^2beta^2-8t^4beta+4t^6 + t^4-2t^2beta+beta^2 = 1 #
# :. (4t^2+1)beta^2-2t^2(4t^2+1)beta +t^4(4t^2+1)=1 #
# :. beta^2-2t^2beta +t^4=1/(4t^2+1) #
# :. beta^2-2t^2beta +t^4-1/(4t^2+1) =0#

We can solve this quadratic in #beta# by completing the square to get:

# (beta-t^2)^2-t^4 +t^4-1/(4t^2+1) =0#
# :. (beta-t^2)^2 = 1/(4t^2+1)#
# :. beta-t^2 = +-sqrt(1/(4t^2+1))#
# :. beta = t^2 +-sqrt(1/(4t^2+1))#

We have found two solutions because there is one point #AP# distance #1# unit extending inwards,and one point extending outwards. Intuitively we choose #beta=t^2 +sqrt(1/(4t^2+1))#.

Now from before we have:

#2tbeta-2t^3 = t -alpha# :
# :. alpha = t -2tbeta+2t^3 #

Substituting our valu of #beta#, from above, gives us:

# alpha = t -2t(t^2 +sqrt(1/(4t^2+1)))+2t^3#
# alpha = t -2t^3 -2tsqrt(1/(4t^2+1))+2t^3#
# alpha = t -2tsqrt(1/(4t^2+1))#

And so the general coordinates of #P(alpha,beta)# in term of #t# are:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

The loci traced out by #P# as #A# moves along the parabola is shown in the following graph in green:

enter image source here

2

Answer:

#f(x) = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))# with radius of convergence #R=1#.

Explanation:

We have:

#f(x) = int_0^x tln(1-t)dt#

Focus on the function:

#ln(1-x)#

We know that:

# ln (1-x) = -int_0^x (dt)/(1-t)#

where the integrand function is the sum of the geometric series:

#sum_(n=0)^oo t^n = 1/(1-t)#

then we can integrate term by term, and we have:

#ln(1-x) = -int_0^x (sum_(n=0)^oo t^n)dt = -sum_(n=0)^oo int_0^x t^ndt =- sum_(n=0)^oo x^(n+1)/(n+1)#

and mutiplying by x term by term:

#xln(1-x) = - sum_(n=0)^oo x^(n+2)/(n+1)#

We can substitute this expression in the original integral and integrate again term by term:

#f(x) = int_0^x tln(1-t)dt = - int_0^x ( sum_(n=0)^oo t^(n+2)/(n+1))dt = -sum_(n=0)^oo int_0^x t^(n+2)/(n+1)dt = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))#

To determine the radius of convergence we can then use the ratio test:

#abs (a_(n+1)/a_n) = abs (frac (x^(n+4)/((n+4)(n+2))) (x^(n+3)/((n+3)(n+1)))) = abs (x) ((n+3)(n+1))/((n+4)(n+2))#

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)#

and the series is absolutely convergent for #abs(x) <1# which means the radius of convergence is #R=1#