Melody

ok I stand corrected.  Thanks :)

a) 10!

1*2*3*4*5*6*7*8*9*10

1*2*3*4*5*6*7*8*9

one

1*3*4*6*7*8*9

two

1*3 ends in 3

3*4 ends in 2

2*6 ends in 2

2*7 ends in 4

4*8 ends in 2

2*9ends in 8

so the last 3 digits are 800

that is 2 zeros at the end.

b) 14!

10!  ends in 800

11! ends in 800

12! ends in 600

13! ends in 800

14! ends in 200

c) 20!    

15! ends in 000

20! ends in 0000        4 zeros

d) 25!

1*2*3*4*5 has 0ne zero at the end

so  add another zero

25!                           5 zeros

I've made a slight error here.  

I can't edit so I will just explain.

dx=-2y dy

I forgot the negative sign so the answer should all be in brackets with a minus sign out the front.

I might as well simplify the answer while I am at it :)

\(\int\;x^2\;\sqrt{2-x}\;dx\\ =-\left[ \frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}\right]+c\\ =-\frac{2}{3*5*7}\left[ \frac{5*7*4(2-x)^{1.5}}{1}-\frac{3*7*4(2-x)^{2.5}}{1}+\frac{3*5*(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{105}\left[ \frac{140(2-x)^{1.5}}{1}-\frac{84(2-x)^{2.5}}{1}+\frac{15(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{140}\left[ 140(2-x)^{1.5}-84(2-x)^{2.5}+15(2-x)^{3.5}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-84(2-x)+15(2-x)^{2}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+15(4-4x+x^2)\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+60-60x+15x^2\right]+c\\ =\frac{-2(2-x)^{1.5}}{140}\left[ 15x^2+24x+32 \right]+c\\\)

3)

\(\int\frac{x}{\sqrt{x+5}}\;dx\\ let\;\;u=x+5\\ so\;\;x=u-5\\ and\;\;dx=du\\~\\ =\int\;\frac{u-5}{\sqrt u}\;du\\ =\int\;u^{0.5}-5u^{-0.5}\;du\\ =\frac{u^{1.5}}{1.5}-\frac{5u^{0.5}}{0.5}+c\\ =\frac{2u^{1.5}}{3}-\frac{10u^{0.5}}{1}+c\\ =\frac{2u^{1.5}}{3}-\frac{30u^{0.5}}{3}+c\\ =\frac{2u^{0.5}}{3}\left[{u}-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x+5-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x-10\right]+c\\ =\frac{2\sqrt{x+5}\;(x-10)}{3}+c\\ \)

2)      \(\int x(sec^2(2x))\;dx\)

I did this with integration by parts.

\(\boxed{\int v \frac{du}{dx}dx=uv-\int u\frac{dv}{dx}dx}\\~\\ v=x \qquad \frac{du}{dx}=sec^2(2x)\\ \frac{dv}{dx}=1 \qquad u=\frac{tan(2x)}{2} \\~\\ \int\;x(sec^2(2x))\;dx\\ =\frac{xtan(2x)}{2}-\int\;1*\frac{tan(2x)}{2}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int tan(2x)\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int \frac{sin(2x)}{cos(2x)}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\frac{ln(cos(2x)}{-2}+c\\ =\frac{xtan(2x)}{2}+\frac{ln(cos(2x))}{4}+c\\\)

1.     integral of x^2(2-x)^0.5 dx

\(\int\;x^2\sqrt{2-x}\;dx\\~\\ \)

2-x has to be positive so I am going to substitute it for  y^2

\(2-x=y^2\\ 2-y^2=x\\ x=2-y^2\\ \frac{dx}{dy}=-2y\\ dx=2y\;dy \)

\(\int\;x^2\;\sqrt{2-x}\;dx\\ =\int \;(2-y^2)^2*\sqrt{y^2}\;*2y\;dy\\ =\int \;(4-4y^2+y^4)*y\;*2y\;dy\\ =\int \;2y^2(4-4y^2+y^4)\;dy\\ =\int \;(8y^2-8y^4+2y^6)\;dy\\ =\frac{8y^3}{3}-\frac{8y^5}{5}+\frac{2y^7}{7}+c\\~\\ =\frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}+c\\~\\\)

I haven't had much practice with these but that looks alright :)

I think so anyway.

Don't be Lazy again like this MWizzard.  I wasted a lot of time today puzzling over your incomplete question !!

Is that a part of the question or what? 

 If it is then the question appears to be relatively trivial.

Are you sure that this question is presented properly and nothing is missing?