Question

Evaluate this indefinite integral (world)joy^world-1d(joy)

Answer 1

let be y= world, and joy=x
so we have integral (world)joy^world-1d(joy) = integ(yx^y-1)dx, y is a constant independant of x, so integ(yx^y-1)dx = y . integ(x^y-1)dx
=y .(x^y-1+1 / y-1 +1) + C=(x^y) + C

the answer is joy^world

Answer 2

The answer is
`joy^(world) + C`

Explanation:

Given the indefinite integral
`\int (world)joy^(world-1)d(joy)`

Let world → a

joy → x

Therefore, the integral becomes
`\int ax^(a-1)dx`

=
`a(x^(a-1+1) )/(a-1+1) + C`

=
`x^(a) + C`

Replacing, a → world

x → joy

Hence,
`\int (world)joy^(world-1)d(joy)` =
`joy^(world) + C`