IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

harshiiit

integrate IxIpower n for n>0

i tried to use IxI =x signum x but i couldnt proceed...........

robinphilip1989

look if ur looking for indefinite integral
let n=2m+1

|x|^n=|x|^2m+1

then for x<0
|x|^2m+1= -x^2m+1 then integrate
for x>0
|x|^2m+1 = x^2m+1 then integrate

similarly for n=2k
|x|^2k = x^2k u can check with k=1
integrate and back substitute the value of m and k

magiclko

case 1 :if n = 2m,

|x|ⁿ= xⁿ = x^2m

then

xⁿ = xⁿ⁺¹/ n+1

case 2 : if n= 2m +1

xⁿ = -

xⁿ if x < 0

= -xⁿ⁺¹/ n+1

xⁿ =

xⁿif x > 0

= xⁿ⁺¹/ n+1

harshiiit

can we do it like this

_{[ ]}

^{[ ]} xpowern signumx^n

and then this will come out to be x^n+1/n+1 signumx^n

and then use IxI^n=x^n signumx^n

the ans will come out to be x IxI^n/n+1 is it ok....................

but the problem is....................

the given ans is x^nIxI/n+1

harshiiit

hey mansi and robin ,
i didnt see ur post wat u have done is fine but is my method incorrect...................

plz reply soon...........................

harshiiit

plz reply yaar............................

10904him

Thats too confusing at the moment for me

harshiiit

nitin and experts plz help..........

nitin62225

soln posted by magiclko is correct and i agree wid it!

just to generalise the two cases mentioned by magiclko,

u can write combining both into one by x^n.IxI/(n+1)

ok ....

soln:

_{[ ]}

^{[ ]} IxI^n dx=sgn(x)

x^n dx= sgn(x).x .x^n/(n+1)=IxIx^n/(n+1)

y did u use sgn(x^n) the value (+ve,or -ve ,not considering the const.) will depend on sign of x and not x^n

like say integral(IxI^2) for x<0 ,is -x^3/3+c and not X^3/3+c

harshiiit

thank u nitin
plz answer to my mechanics ques also

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