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 #1
avatar+18715 
+2
heureka Nov 21, 2017
 #7
avatar+18715 
+2

find the sum of n terms of the series:

 

A) 1+ 2x + 3x^2 + 4x^3 +.....

like Euler:

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1+ 2x + 3x^2 + 4x^3 +\ldots +n x^{n-1}} \quad & | \quad \cdot dx \\ s_n dx &=& dx+ 2xdx + 3x^2dx + 4x^3dx +\ldots +nx^{n-1}dx \quad & | \quad \int{} \\ \int{s_n dx} &=& x+ 2\frac{x^2}{2} + 3\frac{x^3}{3} + 4\frac{x^4}{4} +\ldots +n\frac{x^n}{n} \\ \int{s_n dx} &=& x+ x^2 + x^3 + x^4 +\ldots + x^n = \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx} &=& \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx}&=& \frac{x-x^{n+1}} {1-x} = [x-x^{n+1}][(1-x)^{-1}] \quad & | \quad \text{derivate} \\ s_n &=& [1-(n+1)x^n](1-x)^{-1}+ (x-x^{n+1})(-1)(1-x)^{-2}(-1) \\ s_n &=& \frac{1-(n+1)x^n} {1-x} + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-(n+1)x^n} {1-x} \left(\frac{1-x}{1-x}\right) + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{[1-(n+1)x^n](1-x)+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-x-(n+1)x^n+(n+1)x^{n+1}+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+(n+1)x^{n+1} -x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+[-1+(n+1)]x^{n+1} } {(1-x)^2} \\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{1 -(n+1)x^n+n x^{n+1} } {(1-x)^2} } \\ \hline \end{array}\)

 

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1\cdot 2^2 + 2\cdot3^2 + 3\cdot4^2 +\ldots +n \cdot (n+1)^2 } \\ s_n &=& \sum \limits_{k=1}^{n} k \cdot (k+1)^2 \\ &=& \sum \limits_{k=1}^{n} k \cdot (k^2+2k+1) \\ &=& \sum \limits_{k=1}^{n} (k^3+2k^2+k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + \sum \limits_{k=1}^{n} (2k^2) + \sum \limits_{k=1}^{n} (k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + 2\sum \limits_{k=1}^{n} (k^2) + \sum \limits_{k=1}^{n} (k) \\ && \begin{array}{|rcll|} \hline \sum \limits_{k=1}^{n} (k^3) &=& 1^3+2^3+3^3+ \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\ \sum \limits_{k=1}^{n} (k^2) &=& 1^2+2^2+3^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum \limits_{k=1}^{n} (k) &=& 1+2+3+ \ldots + n = \frac{n(n+1)}{2} \\ \hline \end{array} \\ &=& \left(\frac{n(n+1)}{2}\right)^2 + 2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ &=& \left(\frac{n(n+1)}{2}\right)\left( \frac{n(n+1)}{2} + \frac23\cdot(2n+1) +1\right) \\ &=& \left(\frac{n(n+1)}{2}\right)\left(\frac{3n(n+1)+4(2n+1)+6}{6}\right) \\ &=& \left(\frac{n(n+1)}{12}\right) \Big( 3n(n+1)+4(2n+1)+6 \Big) \\ &=& \frac{n(n+1)[3n(n+1)+4(2n+1)+6]}{12} \\ &=& \frac{n(n+1)(3n^2+3n+8n+4+6)}{12} \\ &=& \frac{n(n+1)(3n^2+11n+10)}{12} \\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{n(n+1)(n+2)(3n+5)}{12}} \\ \hline \end{array}\)

 

laugh

heureka Nov 13, 2017
 #7
avatar+18715 
+2

With gamma and the beta function

 

Formula: \(\displaystyle \int \limits_{0}^{\pi/2} \cos^{2u-1}(x)\cdot \sin^{2v-1}(x) \;dx = \frac12 \cdot B(u,v), \qquad \text{Re } u>0, \quad \text{Re } v>0\)
Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}, \qquad \text{Re } u>0, \quad \text{Re } v>0\)

 

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

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 15 & 2v-1 = 0 \\ 2u = 16 & 2v = 1\\ u = 8 & v = \frac12 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \cos^{15}(x)\cdot \;dx &=& \frac12 \cdot B(8,\frac12) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \Gamma(\frac12)}{\Gamma(8+\frac12)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(8) \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \Gamma(8) = 7! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\Gamma(8+\frac12)} \quad & | \quad \displaystyle \Gamma(8+\frac12) = \frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! \cdot \sqrt{\pi} }{\frac{(2\cdot 8)!}{8!4^8}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{7! 8! 4^8}{16!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 4^8}{ 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14\cdot 15\cdot 16} \\\\ &=& \displaystyle \dfrac{165150720 }{518918400 } \\\\ &=& \displaystyle \dfrac{2048\cdot 80640 }{6435\cdot 80640 } \\\\ &=& \displaystyle \dfrac{2048 }{6435 } \\ \end{array}\)

 

 

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

 

\(\begin{array}{rr} \begin{array}{rr} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx = \frac12 \cdot B(u,v) \\ \end{array}\\ \begin{array}{r|r} 2u-1 = 0 & 2v-1 = 13 \\ 2u = 1 & 2v = 14 \\ u = \frac12 & v = 7 \\ \end{array} \end{array}\)

\(\begin{array}{rcll} \displaystyle \int \limits_{0}^{\pi/2} \sin^{13}(x)\;dx &=& \frac12 \cdot B(\frac12,7) \\ &=& \displaystyle \frac12 \cdot \dfrac{\Gamma(\frac12) \cdot \Gamma(7)}{\Gamma(\frac12+7)} \quad & | \quad \Gamma(\frac12) = \sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot \Gamma(7)}{\Gamma(7+\frac12)} \quad & | \quad \Gamma(7) = 6! \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\Gamma(7+\frac12)} \quad & | \quad \displaystyle \Gamma(7+\frac12) = \frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{\sqrt{\pi} \cdot 6!}{\frac{(2\cdot 7)!}{7!4^7}\sqrt{\pi}} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{6! 7! 4^7}{14!} \\\\ &=& \displaystyle \frac12 \cdot \dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \cdot 4^7}{ 8\cdot 9 \cdot 10 \cdot 11\cdot 12 \cdot 13 \cdot 14} \\\\ &=& \displaystyle \dfrac{5898240 }{17297280 } \\\\ &=& \displaystyle \dfrac{1024\cdot 5760 }{3003\cdot 5760 } \\\\ &=& \displaystyle \dfrac{1024 }{3003 } \\ \end{array}\)

 

laugh

heureka Nov 9, 2017
 #3
avatar+18715 
+2

What is the scale factor used to create the dilation?
The smaller triangle is a pre-image of the bigger triangle. The center of dilation is (2, -1).
What is the scale factor used to create the dilation?

 

Let \(\vec{A} = \binom{-1}{-1} \) before dilation
Let \(\vec{A'} = \binom{8}{-1}\) after dilation
Let \(\vec{C} = \binom{2}{-1}\) the center of dilation
Let \(\lambda \) is the scale factor used to create the dilation

 

Formula for dilation with vector A:
 \(\lambda = -2\)

\(\begin{array}{|rcll|} \hline \vec{A'} &=& (\vec{A}-\vec{C})\cdot \lambda + \vec{C} \quad & | \quad \lambda = -2 \\ \binom{8}{-1} &\overset{?}{=}& \Big(\binom{-1}{-1}-\binom{2}{-1} \Big)\cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-1-2}{-1-(-1)} \cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3}{0} \cdot (-2) + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3\cdot (-2)}{0\cdot (-2)} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{6}{0} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{6+2}{0-1} \\ &\overset{!}{=}& \binom{8}{-1}~ \checkmark \\ \hline \end{array}\)

 

\( \lambda = 2\)
\(\begin{array}{|rcll|} \hline \vec{A'} &=& (\vec{A}-\vec{C})\cdot \lambda + \vec{C} \quad & | \quad \lambda = 2 \\ \binom{8}{-1} &\overset{?}{=}& \Big(\binom{-1}{-1}-\binom{2}{-1} \Big)\cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-1-2}{-1-(-1)} \cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3}{0} \cdot 2 + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-3\cdot 2}{0\cdot 2} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-6}{0} + \binom{2}{-1} \\ &\overset{?}{=}& \binom{-6+2}{0-1} \\ & \ne & \binom{-4}{-1} \\ \hline \end{array}\)

 

 

laugh

heureka Nov 7, 2017
 #3
avatar+18715 
+1

The values of the four variables a, b, c, and d are 9, 11, 13, and 15, though not necessarily in that order.

What is the number of possible values of the expression ab+bc+cd+da?

 

Let a = 9
Let b = 11
Let c = 13
Let d = 15


\(\small{ \begin{array}{|r|c|r|r|r|r|rcl|} \hline & \text{All permutations} & & & & & \\ & \text{of a,b,c,d} & A & B & C & D & AB+BC+CD+DA &=& (B+D)(A+C) \\ \hline 1. & abcd & 9 & 11 & 13 & 15 & (11+15)(9+13) = 26*22 &=& 572 \\ 2. & abdc & 9 & 11 & 15 & 13 & (11+13)(9+15) = 24*24 &=& \qquad 576 \\ 3. & acbd & 9 & 13 & 11 & 15 & (13+15)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 4. & acdb & 9 & 13 & 15 & 11 & (13+11)(9+15) = 24*24 &=& \qquad 576 \\ 5. & adcb & 9 & 15 & 13 & 11 & (15+11)(9+13) = 26*22 &=& 572 \\ 6. & adbc & 9 & 15 & 11 & 13 & (15+13)(9+11) = 28*20 &=& \qquad \qquad 560 \\ 7. & bacd & 11 & 9 & 13 & 15 & (9+15)(11+13) = 24*24 &=& \qquad 576 \\ 8. & badc & 11 & 9 & 15 & 13 & (9+13)(11+15) = 22*26 &=& 572 \\ 9. & bcad & 11 & 13 & 9 & 15 & (13+15)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 10. & bcda & 11 & 13 & 15 & 9 & (13+9)(11+15) = 22*26 &=& 572 \\ 11. & bdca & 11 & 15 & 13 & 9 & (15+9)(11+13) = 24*24 &=& \qquad 576 \\ 12. & bdac & 11 & 15 & 9 & 13 & (15+13)(11+9) = 28*20 &=& \qquad \qquad 560 \\ 13. & cbad & 13 & 11 & 9 & 15 & (11+15)(13+9) = 26*22 &=& 572 \\ 14. & cbda & 13 & 11 & 15 & 9 & (11+9)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 15. & cabd & 13 & 9 & 11 & 15 & (9+15)(13+11) = 24*24 &=& \qquad 576 \\ 16. & cadb & 13 & 9 & 15 & 11 & (9+11)(13+15) = 20*28 &=& \qquad \qquad 560 \\ 17. & cdab & 13 & 15 & 9 & 11 & (15+11)(13+9) = 26*22 &=& 572 \\ 18. & cdba & 13 & 15 & 11 & 9 & (15+9)(13+11) = 24*24 &=& \qquad 576 \\ 19. & dbca & 15 & 11 & 13 & 9 & (11+9)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 20. & dbac & 15 & 11 & 9 & 13 & (11+13)(15+9) = 24*24 &=& \qquad 576 \\ 21. & dcba & 15 & 13 & 11 & 9 & (13+9)(15+11) = 22*26 &=& 572 \\ 22. & dcab & 15 & 13 & 9 & 11 & (13+11)(15+9) = 24*24 &=& \qquad 576 \\ 23. & dacb & 15 & 9 & 13 & 11 & (9+11)(15+13) = 20*28 &=& \qquad \qquad 560 \\ 24. & dabc & 15 & 9 & 11 & 13 & (9+13)(15+11) = 22*26 &=& 572 \\ \hline \end{array} }\)

 

The number of possible values of the expression ab+bc+cd+da=(b+d)(a+c) is 3

The values are 560, 572, and 576

 

laugh

heureka Nov 7, 2017
 #5
avatar+18715 
+1

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

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

 

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

 

laugh

heureka Nov 6, 2017
 #4
avatar+18715 
+2

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

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

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

 

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

 

laugh

heureka Nov 6, 2017
 #2
avatar+18715 
+2

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 \(v = \sqrt{0} = 0\)

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

 

laugh

heureka Nov 6, 2017