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Answer:

# 4sqrt(3)pia^3 #

Explanation:

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

enter image source here

The red circle has radius #2a#, hence its equation is:

# x^2+y^2=(2a)^2 => x^2+y^2=4a^2#

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius #2a#. I will use the standard volume formula #V=4/3pir^3#

# :. V_("sphere") = 4/3 pi (2a)^3 #
# " " = 4/3 pi 8a^3 #
# " " = 32/3 pi a^3 #

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about #Oy#, the shell formula is:

# V = 2pi \ int_(alpha)^(beta) \ xf(x) \ dx #

Also note that we require twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bore is given by:

# V_("bore") = 2pi \ int_(0)^(a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(0)^(a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=0), (x=a) :} => { (u=4a^2), (u=3a^2) :}#

And so:

# V_("bore") = -2pi \ int_(4a^2)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ int_(3a^2)^(4a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(3a^2)^(4a^2) #
# " " = (4pi)/3 \ ((4a^2)^(3/2) - (3a^2)^(3/2)) #
# " " = 4/3pi \ (8a^3-3sqrt(3)a^3) #
# " " = 32/3pi a^3-4sqrt(3)pia^3 #

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #
# " " = 32/3 pi a^3 - (32/3pi a^3-4sqrt(3)pia^3)#
# " " = 4sqrt(3)pia^3#

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about #Oy#, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bead is given by:

# V_("total") = 2pi \ int_(a)^(2a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(a)^(2a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=a), (x=2a) :} => { (u=3a^2), (u=0) :}#

And so:

# V_("total") = -2pi \ int_(3a^2)^(0) \sqrt(u) \ du #
# " " = 2pi \ int_(0)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(0)^(3a^2) #
# " " = (4pi)/3 \ ((3a^2)^(3/2) - 0) #
# " " = (4pi)/3 \ (3sqrt(3)a^3 )#
# " " = 4sqrt(3)pia^3 #, as above

Answer:

# int \ (x-2)/(x(x^2-4x+5)^2) \ dx = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c #

Explanation:

Let us denote the required integral by #I#

# I = int \ (x-2)/(x(x^2-4x+5)^2) \ dx #

Partial Fraction Decomposition

We can decompose the integrand using partial fractions, the partial fraction decomposition will be of the form:

# (x-2)/(x(x^2-4x+5)^2) = A/x + (Bx+C)/(x^2-4x+5) + (Dx+E)/(x^2-4x+5)^2 #
# " " = {A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x}/{x(x^2-4x+5)^2} #

Leading to:

# x-2 = A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x #

# :. x-2 = A(x^4-8x^3+26x^2-40x+25) + (Bx+C)(x^3-4x^2+5x) + (Dx+E)x #

We can use various methods to find our unknown constants:

# "Put " x=0 => -2=A*5^2 #
# :. A = -2/25 #

Compare coefficients:

# "Coeff"(x^4) => 0 = A +B #
# :. B = 2/25 #

# "Coeff"(x^3) => 0 = -8A+C-4B#
# :. 0=16/25+C-8/25 => C=-8/25#

# "Coeff"(x^2) => 0 = 26A+5B-4C+D#
# :. 0=-52/25+10/25+32/25+D => D=10/25=2/5#

# "Coeff"(x^1) => 1 = -40A+5C+E#
# :. 1=80/25-40/25+E => E=-3/5#

Hence we can the integral as:

# I = int \ (-2/25)/x + (2/25x-8/25)/(x^2-4x+5) + (2/5x-3/5)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/35 \ int \ 1/x dx + 2/25 int (x-4)/(x^2-4x+5) \ dx+ 1/5 int (2x-3)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

Where:

# I_1 = int \ 1/x dx + 2/25 \ dx#
# I_2 = int (x-4)/(x^2-4x+5) \ dx #
# I_3 = int (2x-3)/(x^2-4x+5)^2 \ dx#

Now let us take each of these three separate integrals in turn:

Integral 1: #I_1#

The first integral, #I_1#, we can just evaluate directly:

# I_1 = int \ 1/x \ dx #
# \ \ \ = ln|x| #

Integral 2: #I_2#

The second integral, #I_2#, we manipulate by completing the square:

# I_2 = int \ (x-4)/(x^2-4x+5) \ dx #
# \ \ \ = int \ (x-4)/((x-2)^2-2^2+5) \ dx#
# \ \ \ = int \ (x-4)/((x-2)^2+1) \ dx#

Let #u=x-2 => (du)/dx = 1 #, and #x-4=u-2#, then substituting gives

# I_2 = int \ (u-2)/(u^2+1) \ du#
# \ \ \ = int \ u/(u^2+1) - 2/(u^2+1)\ du#
# \ \ \ = 1/2int \ (2u)/(u^2+1) \ du - 2 int \ 1/(u^2+1)\ du#
# \ \ \ = 1/2ln|u^2+1| - 2 arctanu#

Restoring the substitution we get:

# I_2 = 1/2ln|x^2-4x+5| - 2 arctan(x-2)#

Integral 3: #I_3#

The third integral, #I_3# we continue as earlier by completing the square to get:

# I_3 = int \ (2x-3)/(x^2-4x+5)^2 \ dx #
# \ \ \ = int (2x-3)/((x-2)^2+1)^2 \ dx #

Let #u=x-2 => (du)/dx = 1 #, and #2x-3=2u+1#, then substituting gives

# I_3 = int (2u+1)/(u^2+1)^2 \ du #
# \ \ \ = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #

Consider the integral:

# int (2u)/(u^2+1)^2 \ du #

Let #w=u^2+1 => (dw)/(du)=2u#, so we can substitute to get:

# int (2u)/(u^2+1)^2 \ du = int \ 1/w^2 \ dw #
# " " = -1/w #

Restoring the substitutions we get:

# int (2u)/(u^2+1)^2 \ du = -1/(u^2+1) #
# " " = -1/((x-2)^2+1) #
# " " = -1/(x^2-4x+5) #

And now we consider:

# int 1/(u^2+1)^2 \ du #

Let #u=tan theta => (du)/(d theta) = sec^2 theta #, then substituting gives

# int 1/(u^2+1)^2 \ du = int \ 1/(tan^2theta+1)^2 \ sec^2 theta \ d theta #
# " " = int \ 1/(sec^2theta)^2 \ sec^2 theta \ d theta #
# " " = int \ 1/(sec^2theta) \ d theta #
# " " = int \ cos^2theta \ d theta #
# " " = int \ 1/2(1+cos2theta) \ d theta #
# " " = 1/2 \ int \ 1+cos2theta \ d theta #
# " " = 1/2 (theta + 1/2 sin2theta) #
# " " = 1/2 theta + 1/2 sintheta costheta#

Now, #u=tan theta => u=sintheta/costheta #

# :. sinthetacostheta = cos^2theta * u #
# " " =1/sec^2theta * u #
# " " =1/(1+tan^2theta) * u #
# " " =u/(u^2+1) #

Restoring the substitutions we get:

# int 1/(u^2+1)^2 \ du = 1/2 arctanu+ 1/2 u/(u^2+1)#
# " " = 1/2 arctan(x-2) + 1/2 (x-2)/((x-2)^2+1)#
# " " = 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)#

Hence, we can write the third integral as:

# I_3 = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #
# \ \ \ = -1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)#

Combine Results

Now we return back to our original partial fraction result from earlier, and replace the three integrals with the above results:

# I = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

# \ \ = -2/25 {ln|x|} + 2/25 {1/2ln|x^2-4x+5| - 2 arctan(x-2)} + 1/5 {-1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)} + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 4/25 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 arctan(x-2) + 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-2-2)/(10(x^2-4x+5)) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c #

Answer:

#int_1^2(24lnx-6x^2lnx)dx#
#=32ln2 - 58/3#

Explanation:

First set the equations equal to each other, and find the intersection points, then find the area between the curves.

#color(red)(y=6x^2lnx)#
#color(blue)(y=24lnx)#

Desmos

#6x^2lnx=24lnx#
#0=24lnx-6x^2lnx#
#0=(-6)(lnx)(x^2-4)#
#x=1,2#

By setting the equations equal to each other, I found that the functions intersect at #x=1# and #x=2#:

#"Area"=int_1^2(24lnx-6x^2lnx)dx #

Focus on the unbounded (indefinite) integral to solve:
#int(24lnx-6x^2lnx)dx#
#=-6 int (lnx)(x^2-4) dx#

Use integration by parts: #int color(blue)(u) color(green)(dv)= vu - int v du#
#((u=color(blue)(lnx)),(du=1/x dx))((v=1/3x^3-4x),(dv=color(green)(x^2-4)))#

#=-6[(lnx)(1/3x^3-4x)- int(1/x)(1/3x^3-4x)dx]#

#=-6[(lnx)(1/3x^2-4x)-int(1/3x^2-4)dx]#

#=-6[(lnx)(1/3x^2-4x)-(1/9x^3-4x)]#

#=-6(lnx)(1/3x^3-4x)+2/3x^3-24x#

Now go back to the definite integral:
#int_1^2(24lnx-6x^2lnx)dx#
#=[-6(lnx)(1/3x^3-4x)+2/3x^3-24x]_1^2#

#=-6ln2(8/3-8) + 16/3 - 48 - 2/3 + 24#

#= 32ln2- 58/3#

Answer:

Under some conditions, which are given below, letter series converges.

Explanation:

It dependes on properties of #a_n#. There is a more general theorem about this problem:

Theorem: Let #a_n# be a bounded sequnce and #sum b_n# be a convergent series with all #b_n>0#. Then #sum a_n b_n# absolutely converges, means converges.

For our purpose, your hypothesis to be true, we need the followings to be true:

  • For all #a_n>0#,
  • #a_n# sequence is bounded (Let's say #|a_n|<\alpha#),
  • #\sum a_n# be convergent (Let's say its equal to #a#).

If these two items are true, then your statement would be true. Moreover we can find an upper limit for #sum (a_n)^k#, of coarse for positive finite integers #k#.

Let's start with #k=2#:

#sum a_n^2=sum a_n a_n <=sum |a_na_n|<=alpha a#

#sum a_n^2<=alpha a#,

which means #sum a_n^2# converges. Let's pass to #k=3#:

#sum a_n^3=sum a_n a_n^2 <=sum |a_na_n^2|<=alpha sum |a_n^2|<=alpha (alpha a)=alpha^2 a#

#sum a_n^3<=alpha^2 a#

If we continue in this respect we can find by induction (!),

#sum (a_n)^k<=alpha^(k-1) a#.

Answer:

# 2^(n-2)e^(2x){4x^2+4(n-3)x+(n^2-7n+8)}.#

Explanation:

We will solve this Problem, using the following Leibnitz Theorem

(LT) for #n^(th)# Derivative of Product of two functions #u and v.#

Denoting the #n^(th)# der. of fun. #u# by #u_n,# we have,

#(uv)_n=u_nv+""_nC_1u_(n-1)v_1+""_nC_2u_(n-2)v_2+...+u v_n.#

We take, #u=e^(2x) rArr u_n=2^n*e^(2x), and, v=x^2-3x+2.#

#:. v_1=2x-3, v_2=2, &, v_3=v_4=...=v_n=0, n>=3.#

#:. (d^n)/dx^n{(x^2-3x+2)e^(2x)}=(2^n*e^(2x))(x^2-3x+2)#
#+n*(2^(n-1)e^(2x))(2x-3)+n/2*(n-1)*(2^(n-2)e^(2x))*2,#

#=2^n*e^(2x)(x^2-3x+2)+2^(n-1)e^(2x)(2nx-3n)+2^(n-2)e^(2x)(n^2-n),#

#=2^(n-2)e^(2x){4(x^2-3x+2)+2(2nx-3n)+(n^2-n)},#

#=2^(n-2)e^(2x){4x^2+4(n-3)x+(n^2-7n+8)}.#

Enjoy Maths.!

I'm guessing the question refers to the #x# coordinates where #f# changes direction, and that #a xx b = -1/7#:

#a = -sqrt(1/7)#, local minimum

#b = sqrt((1/7)#, local maximum

#f(x) = (6x)/(1 + 7x^2)#:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

#(df)/(dx) = -(6(7x^2 - 1))/(1 + 7x^2)^2#:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

#(d^2f)/(dx^2) = (84x (7x^2 - 3))/(1 + 7 x^2)^3#:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}


DISCLAIMER: LONG ANSWER!

Well, first of all, I think the question has a funky typo. It should say, "changes from decreasing to increasing at the point #a# . . . "

Whenever a function changes from increasing to decreasing or vice versa, it exhibits either a local maximum or minimum.

For example:

  • #y = x^2# has one local minimum at #x = 0#, since the graph has only an upwards concavity.

graph{x^2 [-2.464, 2.464, -2, 2]}

  • #y = sinx# from #0# to #2pi# changes from increasing to decreasing at #x = pi/2#, and changes from decreasing to increasing at #x = (3pi)/2#.

graph{sinx [-0.05, 6.25, -1.2, 1.01]}

For your graph, the first derivative with respect to #x#, denoted #(df)/(dx)#, shows you whether the function increases, decreases, or neither at a certain #x#.

Here, we have (using the product rule):

#(df)/(dx) = 6x cdot -1/(1 + 7x^2)^2 cdot 14x + 1/(1 + 7x^2) cdot 6#

#= (-6(14x^2))/(1 + 7x^2)^2 - (-6(1 + 7x^2))/(1 + 7x^2)^2#

#= (-6(14x^2 - (1 + 7x^2)))/(1 + 7x^2)^2#

#= -(6(7x^2 - 1))/(1 + 7x^2)^2#

So, you took the derivative correctly. Now, to find the points #a# and #b# where the function has a local maximum or minimum, find the zeroes of this derivative (i.e. find when #(df)/(dx) = 0#, the function changes direction).

#0 = -(cancel(6)^(ne 0)(7x^2 - 1))/cancel((1 + 7x^2)^2)^(ne 0)#

The denominator can never equal #0# because it's an upwards-shifted quadratic, so we don't have to worry about it.

#=> 7x^2 = 1#

So, your critical x values (for your local maxima and/or minima) are at

#=> color(green)(x) = pmsqrt(1/7) ~~ color(green)(pm 0.3780)#

Now that you know the #x# coordinates, you'll need the #y# coordinates, so plug in your #x# values back into the ORIGINAL function:

#color(green)(f(pm sqrt(1/7))) = (6(pmsqrt(1/7)))/(1 + 7(pmsqrt(1/7))^2)#

#= (pm6/(sqrt7))/(1 + 7 cdot 1/7) = pm3/(sqrt7)#

#~~ color(green)(pm 1.134)#

Right now, all you know is where the points #a# and #b# are, but not which is which.

The second derivative tells you the concavity; up, down, or an inflection point. By the question wording, one is concave up and the other is concave down.

This derivative would be somewhat ugly

#(d^2f)/(dx^2) = -d/(dx)[(42x^2 - 6)/(1 + 7x^2)^2]#,

but you should get:

#= (84x (7x^2 - 3))/(1 + 7 x^2)^3#

At the same critical #x# values you got earlier, evaluate this second derivative.

  • If it is positive, the function is concave up and is at a minimum.
  • If it is negative, the function is concave down and is at a maximum.
  • Otherwise, the function is at an inflection point.

At #ul(x = -sqrt(1/7))#:

#ul((d^2f)/(dx^2)) = (84(-sqrt(1/7)) (7(-sqrt(1/7))^2 - 3))/(1 + 7 (-sqrt(1/7))^2)^3#

#= ((-84/sqrt7) (-2))/(2)^3 " "ul(> 0)#

At #ul(x = +sqrt(1/7))#, we similarly get:

#ul((d^2f)/(dx^2) < 0)#

So, the left-hand critical point is a minimum and the right-hand critical point is a maximum.

#color(blue)((-sqrt(1/7), -3/sqrt7) ~~ (-0.3780, -1.134))# (minimum)

#color(blue)((sqrt(1/7), 3/sqrt7) ~~ (0.3780, 1.134))# (maximum)

And we can check by graphing. Here is the function, followed by its first derivative, and then its second derivative.

#f(x) = (6x)/(1 + 7x^2)#:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

#(df)/(dx) = -(6(7x^2 - 1))/(1 + 7x^2)^2#:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

#(d^2f)/(dx^2) = (84x (7x^2 - 3))/(1 + 7 x^2)^3#:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]

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