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yea it was amazing

I got to spend the night at my Aunts

volume  =  area of base * height

volume  =  pi * radius² * height            radius  =  12/2  and  height = 10

volume  =  pi * (12/2)² * 10

volume  =  pi * 6² * 10   =   pi * 36 * 10   =   360pi  in³   ≈   1130.973  in³

surface area  =  circumference of base * height  +  2 * area of base

surface area  =  pi * diameter *height  +  2 * pi * radius²

surface area  =  pi * 12 * 10  +  2 * pi * 6²

surface area  =  120pi  +  2 * pi * 36

surface area  =  120pi + 72pi   =   192pi   in²   ≈   603.186  in²

thanks!!

f(x)  =  \(\frac{3x-7}{x+1}\)                    Let's say...

y  =  \(\frac{3x-7}{x+1}\)                               Now we want to solve for  x .

                                                Multiply both sides of the equation by  x + 1 .

y(x + 1)  =  3x - 7

                                                Distribute the  y .

yx + y  =  3x - 7

                                                Subtract  3x  from both sides.

yx - 3x +  y  =  -7

                                                Factor  x  out of the first two terms.

x(y - 3) + y  =  -7

                                                Subtract  y  from both sides.

x(y - 3)  =  -y - 7

                                                Divide both sides by  y - 3 .

x  =  \(\frac{-y-7}{y-3}\)

                                      Now to get the inverse, swap  x  and  y .

y  =  \(\frac{-x-7}{x-3}\)                    This is the inverse function.

f^-1(x)  =  \(\frac{-x-7}{x-3}\)

f(x)  =  2x + 7

g(x)  =  3x + c

f( g(x) )  =  f( 3x + c )  =  2( 3x + c ) + 7

g( f(x) )  =  g( 2x + 7 )  =  3( 2x + 7 ) + c

And we want to know what  c  is when  f( g(x)  )  =  g( f(x) )  . That is... when

2( 3x + c ) + 7  =  3( 2x + 7 ) + c           Distribute the numbers outside the parenthesees.

6x + 2c + 7  =  6x + 21 + c                   Subtract  6x  from both sides of the equation.

2c + 7  =  21 + c             Subtract  c  from both sides of the equation.

c  + 7  =  21                    Subtract  7  from both sides of the equation.

c  =  14

Kind of. You got the 5 right and the 1/6 but it's 1/x+1. It's together

y  =  -1/2 x - 3

The point where the line crosses the  y-axis  is the point where  x = 0 .

So, plug in  0  for  x  and solve for  y .

y  =  -1/2 (0) - 3

                                       And  -1/2 * 0  =  0 .

y  =  0 - 3

y  =  -3

When  x = 0 ,  y = -3 .    That gives us the point  (0, -3) .

Wow thank you so much! This helps a lot! :)

That is correct.

integrals

5) \(\displaystyle \int^{\pi/2}_{0}\sin^{13}x\;dx\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\sin^{12}(x)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x)\Big(\sin^2(x)\Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big(1-\cos^2(x) \Big)^{6}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \sin(x) \Big( \binom{6}{0}-\binom{6}{1}\cos^2(x) \\ && +\binom{6}{2}\cos^4(x)-\binom{6}{3}\cos^6(x)+\binom{6}{4}\cos^8(x) \\ && -\binom{6}{5}\cos^{10}(x)+\binom{6}{6}\cos^{12}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ \end{array}\)

Integration by parts
\(\begin{array}{rclrcl} u&=&\cos^n(x) & v &=& -\cos(x) \\ u'&=&n\cos^{n-1}(x)\sin(x) & v' &=& \sin(x) \\ \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\cos^n(x)}_{u}\underbrace{\sin(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\cos^n(x)}_{u}(\underbrace{-\cos(x)}_{v}) \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{-n \cos^{n-1}(x)\sin(x)}_{u'}(\underbrace{-\cos(x)}_{v})\;dx \\ &=& \Big[ -\cos^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\cos^{n}(x)\sin(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx &=& -\Big[ \cos^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \cos^n(x)\sin(x)\;dx} & \mathbf{=} & \mathbf{\frac{-\Big[\cos^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array} }\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\cos^0(x)\sin(x) -6\cos^2(x)\sin(x) \\ && +15\cos^4(x)\sin(x)-20\cos^6(x)\sin(x)+15\cos^8(x)\sin(x) \\ && -6\cos^{10}(x)\sin(x)+1\cos^{12}(x)\sin(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 6\cdot \frac{1}{3}+15\cdot \frac{1}{5} - 20\cdot \frac{1}{7}+15\cdot \frac{1}{9}- 6\cdot \frac{1}{11}+1\cdot \frac{1}{13} \\\\ &=& \mathbf{\dfrac{1024}{3003} } \\ \end{array}\)

integrals

4) \(\displaystyle \int^{\pi/2}_{0}\cos^{15}{x}\;dx\)

without substitution:

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\cos^{14}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x)\Big(\cos^2(x)\Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big(1-\sin^2(x) \Big)^{7}{x}\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \cos(x) \Big( \binom{7}{0}-\binom{7}{1}\sin^2(x) \\ && +\binom{7}{2}\sin^4(x)-\binom{7}{3}\sin^6(x)+\binom{7}{4}\sin^8(x) \\ && -\binom{7}{5}\sin^{10}(x)+\binom{7}{6}\sin^{12}(x)-\binom{7}{7}\sin^{14}(x) \Big)\;dx \\ &=& \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ \end{array}\)

Integration by parts

\(\begin{array}{rclrcl} u&=&\sin^n(x) & v &=& \sin(x) \\ u'&=&n\sin^{n-1}(x)\cos(x) & v' &=& \cos(x) \\ \end{array}\)

\(\begin{array}{|rcll|} \hline && \displaystyle \int \limits_{0}^{\pi/2} \underbrace{\sin^n(x)}_{u}\underbrace{\cos(x)}_{v'}\;dx \\ &=& \Big[ \underbrace{\sin^n(x)}_{u}\underbrace{\sin(x)}_{v} \Big]_{0}^{\pi/2} - \int \limits_{0}^{\pi/2} \underbrace{n \sin^{n-1}(x)\cos(x)}_{u'}\underbrace{\sin(x)}_{v}\;dx \\ &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} - n \int \limits_{0}^{\pi/2}\sin^{n}(x)\cos(x)\;dx \\ \\ (n+1)\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx &=& \Big[ \sin^{n+1}(x)\Big]_{0}^{\pi/2} \\ \mathbf{\displaystyle \int \limits_{0}^{\pi/2} \sin^n(x)\cos(x)\;dx} & \mathbf{=} & \mathbf{\frac{\Big[\sin^{n+1}(x)\Big]_{0}^{\pi/2} } {n+1} = \frac{1}{n+1} }\\ \hline \end{array}\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{\pi/2} \Big( 1\sin^0(x)\cos(x) -7\sin^2(x)\cos(x) \\ && +21\sin^4(x)\cos(x)-35\sin^6(x)\cos(x)+35\sin^8(x)\cos(x) \\ && -21\sin^{10}(x)\cos(x)+7\sin^{12}(x)\cos(x)-\sin^{14}(x)\cos(x) \Big)\;dx \\ &=& 1\cdot\frac{1}{1} - 7\cdot \frac{1}{3}+21\cdot \frac{1}{5} - 35\cdot \frac{1}{7}+35\cdot \frac{1}{9}- 21\cdot \frac{1}{11}+7\cdot \frac{1}{13} - 1\cdot \frac{1}{15} \\\\ &=& \mathbf{\dfrac{2048}{6435} } \\ \end{array}\)

4)

pi/ 2

 ∫     ( cos x)¹⁵   dx      =  

0

pi/ 2

 ∫   cos x   ( cos x)¹⁴   dx      = 

0

pi/ 2

 ∫   cos x   ( cos ²x)⁷   dx      = 

0

pi/ 2

 ∫   cos x  ( 1 - sin² x)⁷   dx       

0

Let   u  =  sin x       →  du   = cos x dx

And  the new limits are :   sin (pi/2)  = 1      and  sin (0)   = 0

1

 ∫  ( 1 - u^2)⁷  du

0

1

∫  -u^14 + 7 u^12 - 21 u^10 + 35 u^8 - 35 u^6 + 21 u^4 - 7 u^2 + 1    du

0

                                                                                                                          1  

 [ - (1 /15)u¹⁵+(7/13)u¹³ - (21/11)u¹¹+(35/9)u⁹ -(35/7)u⁷+(21/5)u⁵- (7/3)u³+ u ]

                                                                                                                          0

 [ -1/15   +  7/13  -  21/11  +  35/9   -  5  + 21/5  - 7/3  + 1 ]   

2048 / 6435 

lim                sin (x)

x → pi          _____

                     x -  pi

Note   sin (pi- x)  =  sin pi cosx  - sinx cospi  =   sin (x)   .....so........

lim            sin ( pi - x)         

x → pi      ________     

                      x - pi           

 1           lim               sin ( pi - x)

__  *     x → pi        __________

-1                                ( pi - x )

Note       lim        sin (pi - x)                      lim          sin (x)

             x→ pi     _______       =           x → 0        _____        =    1

                               pi - x                                            x

So

lim                sin (x)

x → pi          _____        =   (-1) * (1)    =   -1

                     x -  pi

integrals

2) \(\displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx\)

\(\begin{array}{rcll} && \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}-\sqrt{x}}\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt{\sqrt[3]{x}\left(1-\frac{\sqrt{x}}{\sqrt[3]{x}} \right) }\;dx \\ &=& \displaystyle \int \limits_{0}^{1}\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x} }\;dx \\ \end{array} \)

For the integrand \(\sqrt[6]{x}\sqrt{ 1-\sqrt[6]{x}}\), substitute \(u = \sqrt[6]{x}\) or \(u^5 = x^{\frac{5}{6}}\) and

\(\begin{array}{rcll} du &=& \frac16 x^{\frac16-1} \;dx \\ &=& \frac16 x^{-\frac{5}{6}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{ x^{\frac{5}{6}}} \;dx \\ &=& \dfrac16\cdot \dfrac{1}{u^5} \;dx \\ \text{so } dx &=& 6u^5\;du \\ \end{array} \)

This gives a new lower bound \(u = \sqrt[6]{0} = 0\) and upper bound \(u = \sqrt[6]{1} = 1\)

\(\begin{array}{rcll} &=& \displaystyle \int \limits_{0}^{1}u\sqrt{1-u}\cdot 6u^5\;du \\ &=& 6\displaystyle \int \limits_{0}^{1}u^6\sqrt{1-u} \;du \\ \end{array} \)

For the integrand \(u^6\sqrt{1-u}\), substitute \(v = \sqrt{1-u}\)  or \(u = 1-v^2\) and

\(\begin{array}{rcll} dv &=& \dfrac{-1}{2\sqrt{1-u}} \;du \\ &=& \dfrac{-1}{2v} \;du \\ \text{so } du &=& -2v \;dv \\ \end{array}\)

This gives a new lower bound \(v = \sqrt{1} = 1\) and upper bound \(v = \sqrt{0} = 0\)

\(\begin{array}{rcll} &=& 6\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v\cdot (-2)v\;dv \\ &=& -12\displaystyle \int \limits_{1}^{0}(1-v^2)^6 v^2\;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} (1-6v^2+15v^4-20v^6+15v^8-6v^{10}+v^{12} )v^2 \;dv \\ &=& -12 \displaystyle \int \limits_{1}^{0} ( v^2-6v^4+15v^6-20v^8+15v^{10}-6v^{12}+v^{14}) \;dv \\ &=& -12 \cdot \Big[ \frac{v^3}{3}-6\frac{v^5}{5}+15\frac{v^7}{7}-20\frac{v^9}{9}+15\frac{v^{11}}{11}-6\frac{v^{13}}{13}+\frac{v^{15}}{15} \Big]_{1}^{0} \\ &=& 12 \cdot\Big( \frac{1}{3}-\frac{6}{5}+\frac{15}{7}-\frac{20}{9}+\frac{15}{11}-\frac{6}{13}+\frac{1}{15} \Big) \\ &=& 12 \cdot\Big( \frac{675675-6\cdot405405+15\cdot289575-20\cdot225225+15\cdot184275-6\cdot 155925+135135}{2027025} \Big) \\ &=& \dfrac{552960}{2027025} \\ \\ &=& \dfrac{135\cdot 4096}{135\cdot 15015} \\ \\ &=& \dfrac{ 4096}{ 15015} \\ \end{array} \)

lim           √ [2 - x ] - 1

x →1        _________

                     x  - 1

lim           √ [2 - x ] - 1        √ [2 - x ] + 1 

x →1        _________        __________

                     x  - 1              √ [2 - x ] + 1 

lim                2  - x  - 1

x → 1          __________________

                   ( x - 1) (   √ [2 - x ] + 1)

lim

x → 1          1 - x

                   ___________________

                   ( x  - 1)  (   √ [2 - x ] + 1)

lim             - ( x  - 1 )

x → 1       ______________________

                    ( x - 1)  (   √ [2 - x ] + 1)

lim                 -1                                                     -1

x → 1          _________________      =              ____     

                     √ [2 - x ] + 1                                       2

https://www.desmos.com/calculator/ev6xzdsszs

5.1( a )   arcsin [ cos (pi/3) ] - sin [ arcsin (-1/2) ]  =

               arcsin [ 1/2]  -   (-1/2)

                  pi / 6  +  1/2

      (b)  arctan [cos (pi) ] -  arctan [ tan (3pi/ 7) ]  =

             arctan [ -1]    -     3pi/ 7

                - pi/ 4    -   3pi / 7

                 - [ 7 pi  + 12 pi ]  /28

                   - 19pi  / 28

5.2 ( a)  arccos [ cos (pi/5) ] - sin [ arctan (-1) ]  =

                   pi  / 5    -    sin [ - pi / 4 ]

                   pi  / 5   -   [ - 1 / √2] 

                   pi  / 5   +  [ 1 / √2]

      (b )  arctan [ cos (pi/3) ] -  tan [ arctan (3pi / 7) ] =

              arctan [ 1/2 ]  -  3pi / 7 

5.3   (a)  arccos [ cos (pi / 5) ] - sin [ arctan (-1) ]  =

                pi  / 5    -  sin  [ - pi / 4 ] =

                pi  / 5  -  [ - 1 / √2 ]  =

                pi / 5   +  [1 / √2 ]

          (b) same as  5.2 (b)

Thanks Alan,

Another for me to study  

All studied.

Equating coefficients - maybe I will think of it myself next time.   Thanks once again :)

19 + 7 ÷ 2  - 5  =

14  - 7 ÷ 2  =

21/2

[ 19 +  7 ] ÷ 2   - 5 

26 ÷ 2   - 5

13  - 5

8 

[ 19 +  7 [  ÷  [ 2   x  6   +  1 ]

26  /   [ 12 + 1 ]

26 / 13

2