 
( ô MEGA MATHS (INTEGRATION 1) INTEG1 $ ô COPYRIGHT (C) G.LUDINSKI 1985  î… ù   ( *KEY10"OLD|M RUN|M"   2 *KEY 0 "‡" F ë4 P C0=0:C1=0:C2=1:C3=1 d ï19,0,4,0,0,0 n2 S$="                                        " x ï23,1,0;0;0;0; ‚B MI$="‡"+ÀS$,38)+"‡":TP$="…"+Ä38,"‚")+"†":BT$="ƒ"+Ä38,"‚")+"„" ŒÄ Z$="‡‡    b …                               ‡‡and   ‡  f(X) dX =  F(b) - F(a)       ‡‡    a „                               ‡‡where F(b) is the integral of f(X)    ‡‡with b substituted for X" –4 Y$="…                                       ‡ "   ûC1:ûC3+128 ªE ñŠ0,3);S$;" Integration 1  (Mega Maths) ";½(225);" LCL 1985 ";S$ ´ ûC0+128:ûC3 ¾ TE=0:MK=0:TW=0:TS=0:Ñ=0   È W=1:T=1:I$="" Ê çTE/2=¨(TE/2)ŒCC=4 ‹ CC=1 Ì ï19,0,CC,0,0,0 Ò
 ädFB Ü TE=TE+1  æ çTE=1 €T=2 Œï19,1,4,0,0,0 è% ñ:ñQ$:ñ:ñTP$;"‡";ÀS$,38);"‡";BT$ ê çT=2Œñ:ï11 ì! ï11:ï11:ï11:ï9:ï19,1,7,0,0,0 ð
 ò_IN  òB çTE=1 €T=1Œ XP=±:VP=¼:ñŠ0,3);S$;S$;S$:ñŠXP,VP);:ï19,1,7,0,0,0 ôA çI$=½(135)Œñ½(225);½(76);½(67);½(76);½(49);½(57);½(56);½(53) ú î… äDlB  çI$="?"ŒäDbB  î… ù   çI$=""ŒåtZA" ç ”(»(I$)-A)< 0.01 ŒåD|A 6
 åtZA| ûC3:ñ" ";½(224):ûC3† TS=TS+1K Ô1,-10,12,10:Ô1,-10,20,10:Ô1,-10,28,10:Ô1,-10,32,20:Ô1,-10,14,20:åtxAšD çT=1Œñ:ñ:ñTP$;"‡No,";H$;H1$;",try again":ï8:ñ"‡";BT$;:T=2:ådf@¤Y ï19,1,CC,0,0,0:ñ:ñ:ñ:ñ"Sorry,the answer is = ":ñ:ñL$:ñM$:ñN$:ñO$:ñP$:ñ:ï19,1,7,0,0,0® TW=TW+1¸C û128+C3:ûC0:ñŠ0,31);"Do you want more ? (Y/N)               ";Â MK=¨(100*TS/(TS+TW)) Ì R=¦(0):çR=-1ŒådLAÖ û128+C0:ûC3à çR=78Œ×"MENU" ê T=1:Û:ñ:ñ:ådH@ô
 Ýò_INþ *FX15,0
 I$="" I1=¦(0)  I1$=½(I1):çI1<>-1ŒåTpB&h XP=±:YP=¼:ûC2:ñŠ0,1);"Test "+Ã(TE)+"  Score "+Ã(MK)+"%  "+Ã(¨(‘/100))+" secs. ":ûC3:ñŠXP,YP);:åTRB0! çI1=13„©(I$)>37„I1=127ŒåDDB:} XP=±+1:YP=¼:ñI1$;:ûC2:ñŠ0,1);"Test "+Ã(TE)+"  Score "+Ã(MK)+"%"+"  "+Ã(¨(‘/100))+" secs. ":ûC3:ñŠXP,YP);:I$=I$+I1$:åTRBD çI1<>127„I$=""ŒåDXBN# I$=ÀI$,©(I$)-1):ñ½(127);:åTRBX áb
  VP=¼lW û128+C3:ûC1:ñ Š0,29);"             Calculator mode            ";Š0,29):û128+C0:ûC3v
 B$=""€H I$=¾:ñI$;:çI$<>½(127) € I$ <> "="€I$ <> "?"€B$<>"AC"ŒB$=B$+I$:åt@BŠ/ çI$=½(127) € B$<>"" ŒB$=ÀB$,©(B$)-1):åt@B” ç B$="?"„I$="?"ŒåtrBž, çB$="AC"ŒñŠ0,30);S$;Š0,29):B$="":åDvB ¨ ñ (B$);Š0,29):åDvB² ñŠ0,29);S$;S$;Š1,VP);:ò_IN¼ øÆ0 L$="":M$="":N$="":O$="":P$="":H1$="":W=³(7)Ð( F=³(2)+1:E=(F+1)*(³(4)+1)*(-1)^³(2)Ú+ AL=(³(7)+1):BL=³(8)+1:çBL <= AL ŒådZBÜ( E$=Ã(E):F$=Ã(F):BL$=Ã(BL):AL$=Ã(AL)Þ çW<>1ŒåT`CäF Q$=BL$+Y$+E$+"X"+½(11)+F$+½(10)+" dX"+ÀS$,33-©(E$)-©(F$))+AL$+"„"î¥ H$="              n    n+1             ‡‡the integral of X  = X     + constant ‡‡                     ‚‚‚‚             ‡‡                     n+1              "ø  H1$=Z$P A=E*(BL^(F+1))/(F+1)-(E*(AL^(F+1))/(F+1)):IA%=A:ç ”(A-IA%)<0.1 Œ A$=Ã(IA%) & ç ”(A-IA%)>0.9 Œ A$=Ã(IA%+´(A))  ‡ L$=A$:M$=S$+"as the integral is "+E$+"X"+½(11)+F$+"+1"+½(10)+" + c":N$="                   ‚‚‚‚":O$="                    "+F$+"+1"l P$=S$+"subs. "+E$+"x("+BL$+½(11)+F$+"+1"+½(10)+"/("+F$+"+1) - "+AL$+½(11)+F$+"+1"+½(10)+"/("+F$+"+1) )"  çW<>2ŒåD\C*- Q$=BL$+Y$+E$+" dX"+ÀS$,34-©(E$))+AL$+"„"4- H$="the integral of a constant k = kX+c"> H1$=Z$H  A=E*BL-E*AL:A$=Ã(A)R[ L$=A$:M$="as the integral is "+E$+"X + c":N$="substituting,"+E$+"x("+BL$+" - "+AL$+")"\ çW<>3 € W<>4ŒåDbC^o ç³(2)=2Œ BL$="Ž/3":AL$=" 0 ":OB$=BL$:OA$="0":BL=¯/3:AL=0 ‹ AL$="Ž/3":BL$=" 0 ":AL=¯/3:BL=0:OB$="0":OA$=AL$` E=³(9)*2*(-1)^³(2):E$=Ã(E)b çW<>3ŒåtXCf3 Q$=BL$+Y$+E$+"SIN(X) dX"+ÀS$,26-©(E$))+AL$+"„"pU H$="the integral of SIN(X) = -COS(X)+c ‡‡and COS(0)=1 and COS(Ž/3)=1/2         "z H1$=Z$ CB=0.5:ç BL=0 Œ CB=1 CA=0.5:ç AL=0 Œ CA=1 „ A=-E*(CB-CA):òS(A):A$=SS$Ž¬ L$=A$:M$="as the integral is "+Ã(-E)+"COS(X) + c":N$="substituting,"+Ã(-E)+"x(COS("+OB$+") - COS("+OA$+"))":òS(CB):SB$=SS$:òS(CA):O$="= "+Ã(-E)+"x("+SB$+" - "+SS$+")" ˜ çW<>4ŒådJCšA çAL=¯/3Œ AL=¯/2:AL$="Ž/2":OA$=AL$ ‹ BL=¯/2:BL$="Ž/2":OB$=BL$¢3 Q$=BL$+Y$+E$+"COS(X) dX"+ÀS$,26-©(E$))+AL$+"„"¬\ H$="the integral of COS(X) = SIN(X)+c  ‡‡and SIN(0)=0 and SIN(Ž/2)=1           ":H1$=Z$¶$ A=E*(µ(BL)-µ(AL)):òS(A):A$=SS$ À› L$=A$:M$="as the integral is "+E$+"SIN(X) + c":N$="substituting,"+E$+"x(SIN("+OB$+") - SIN("+OA$+"))":òS(µ(AL)):O$="= "+E$+"x("+Ã(µ(BL))+" - "+SS$+")"Ê çW<>5ŒåTPDÔn ç³(2)=1ŒAL=0:AL$=" 0 ":OA$="0":BL=¯/4:BL$="Ž/4":OB$=BL$ ‹ BL=0:BL$=" 0 ":OB$="0":AL=¯/4:AL$="Ž/4":OA$=AL$Þ! E=(³(8)+1)*(-1)^³(2):E$=Ã(E)èF Q$=BL$+Y$+E$+"SEC"+½(11)+"2"+½(10)+"(X) dX"+ÀS$,25-©(E$))+AL$+"„"ò} H$="                2                  ‡‡the integral of SEC (X) = TAN(X)+c    ‡‡and TAN(0)=0 and TAN(Ž/4)=1           "ô H1$=Z$ü# A=E*(·(BL)-·(AL)):òS(A):A$=SS$– L$=A$:M$="as the integral is "+E$+"TAN(X) + c":N$="substituting,"+E$+"x(TAN("+OB$+") - TAN("+OA$+"))":O$="= "+E$+"x("+Ã(·(BL))+" - "+Ã(·(AL))+")"  çW<>6ŒåDVD] ç³(2)=1ŒAL=¡(1):AL$="e":BL=1:BL$="1":LA=1:LB=0  ‹ BL=¡(1):BL$="e":AL=1:AL$="1":LA=0:LB=1$! E=(³(8)+1)*(-1)^³(2):E$=Ã(E)./ Q$=BL$+Y$+E$+"/X dX"+ÀS$,32-©(E$))+AL$+"„"8\ H$="the integral of 1/X = LN(‡X‡)+c    ‡‡and LN(1)=0 and LN(e)=1               ":H1$=Z$B A=E*(LB-LA):òS(A):A$=SS$  L L$=A$:M$="as the integral is "+E$+"LN(ˆXˆ)+c":N$="substituting,"+E$+"x(LN("+BL$+") - LN("+AL$+"))":O$="= "+E$+"x("+Ã(LB)+" - "+Ã(LA)+")"V çW<>7ŒådPG`B ç³(2)=1ŒAL=0:AL$="0":BL=1:BL$="1" ‹ BL=0:BL$="0":AL=1:AL$="1"j! E=(³(8)+1)*(-1)^³(2):E$=Ã(E)tW Q$=BL$+Y$+E$+"e"+½(11)+"X"+½(10)+" dX"+ÀS$,32-©(E$))+AL$+"„           given e=2.7"~­ H$="              X    X               ‡‡the integral of e  = e +c             ‡‡     0                                ‡‡and e =1 and e=2.7                    ":H1$=Z$ ˆ çAL=0ŒA=E*1.7 ‹ A=-1.7*E’~ òS(A):A$=SS$:L$=A$:M$="as the integral is "+E$+"e"+½(11)+"X"+½(10)+"+c":N$="substituting,"+E$+"x(e^"+BL$+" - e^"+AL$+")" œ= çAL=0 ŒO$="= "+E$+"x(2.7 - 1)" ‹ O$="= "+E$+"x(1 - 2.7)"Ð øÚ Ý òS(GG)    ä, IG%=GG:ç ”(GG-IG%)<=.001 Œ SS$=Ã(IG%):áî, D%=100*(GG-IG%):SS$=Ã(IG%)+"."+Ã(”(D%))ø áÿ