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3 years ago

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

Mathematics

2 answers:

Sidana [21]3 years ago

8 0

Answer:

The answer is $joy^{world} + C$

Step-by-step explanation:

Given the indefinite integral $\int (world)joy^{world-1}d(joy)$

Let world → a

joy → x

Therefore, the integral becomes $\int ax^{a-1}dx$

= $a\frac{x^{a-1+1} }{a-1+1} + C$

= $x^{a} + C$

Replacing, a → world

x → joy

Hence, $\int (world)joy^{world-1}d(joy)$ = $joy^{world} + C$

lapo4ka [179]3 years ago

5 0

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

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