backup calcus
TRANSCRIPT
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7/26/2019 Backup calcus
1/2
> a:=exp(x)/2;
:=a1
2ex
>b:= 15*ln(x^2);
:=b 15 ( )ln x2
> fsolve(a=b,x);-1.006112566
> evalf(a=b);
=0.5000000000 ex
15. ( )ln x2
> a=b;
=
1
2ex
15 ( )ln x2
> c:=a=b;
:=c =1
2ex
15 ( )ln x2
> solve(c,x);
( )RootOf e_Z
30 ( )ln _Z2
> fsolve(c,x);
-1.006112566
> d:=b-a;
:=d 15 ( )ln x2
1
2ex
> int(d,x=0..4.5);
23.54188291
> %/2;
11.77094146
> int(d,x=0..u);
+
1
215 ( )ln u
2u 30u
1
2eu
> fsolve(%,u);
5.416292883
> int(d,x=1..4.5);
54.40102383
> e:=int(d,x=1..u)=(int(d,x=1..4.5))/2;
:=e =+ + 301
2e 15 ( )ln u
2u 30u
1
2eu
27.20051192
> fsolve(%,u);
5.472653137
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7/26/2019 Backup calcus
2/2
> (int(d,x=1..4.5))/2
> ;
27.20051192
>m:= int(d,x=1..p)=(int(d,x=1..4.5))/2;
:=m =+ + 301
2e 15 ( )ln p
2p 30p
1
2ep
27.20051192
> fsolve(m,p);
5.472653137
> solve(m,p);Warning, solutions may have been lost
> solve(m,p);Warning, solutions may have been lost
> evalf(m,p);Error, invalid input: evalf expects its 2nd argument, n, to be of type posint, but
received p
>