Featured 5 days ago

We need to split up the integral.

Recall that

so

We are integrating on

Therefore,

# = 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

# = 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

The notation for this technique is

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.)

Featured 5 days ago

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.

Given:

Begin reducing by adding 0 to the numerator in the form of

Break into two fractions:

Factor

The first term becomes

Add 0 to the numerator of the second term in the form of

Break the second term into two fractions:

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

The second term becomes 3:

Partial Fraction Expansion of the last term:

Make B disappear by letting x = -2

Make A disappear by Letting x = 1:

Check

This checks

Returning to the main problem:

Featured 5 days ago

The integral is

This can be rewritten as

The integration by parts formula states that an integral

Use the formula now:

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

This is a known integral that is derived here.

We now reverse the substitutions by drawing a triangle and finding the correct ratios. We know from our original substitution that

Let's return our attention to the whole integral.

Hopefully this helps!

Featured 4 days ago

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

#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

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

Reverse the substitution, since the initial variable wasn't

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

Evaluate using the second fundamental theorem of calculus, which states that

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

#2 + 1/2 - 2#

#1/2#

Hopefully this helps!

Featured yesterday

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

Integration by parts takes the form

#u=lnx#

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

Differentiate

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

Then:

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

Letting

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

#B=-1#

Letting

#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

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

Featured 4 days ago

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

I assume that the second terms include

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

#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

#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

#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!

Featured 2 days ago

First we rearrange the equation of the surface into the form

# 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

# 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

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

Where

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:

and here is the surface with the tangent plane and the normal vector:

Featured yesterday

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

Where

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

# 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) #

Featured 21 hours ago

#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)) #

Our parabola

Differentiating wrt

#dy/dx=2x = 2t #

So the gradient of the tangent at

We can now form the equation of the normal

# y-t^2=-1/(2t)(x-t) #

# :. 2ty-2t^3=-(x-t) #

# :. 2ty-2t^3=+t-x #

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

And we are also told that

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

We can combine these equations to eliminate

# (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-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

Now from before we have:

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

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

Substituting our valu of

# 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 = 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

Featured 21 hours ago

We have:

Focus on the function:

We know that:

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

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

and mutiplying by x term by term:

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

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

and the series is absolutely convergent for