#### To determine

**To evaluate:**

∫02y21+y3dy if it exists

#### Answer

**529**

#### Explanation

**1) Concept:**

By using the substitution rule for definite integrals and fundamental rule of calculus

**2) Theorem and Rule:**

Fundamental theorem of calculus:

If f is continuous on [a, b], then ∫abfxdx=Fb-Fa. Substitution rule for definite integrals:

If g' is continuous on [a, b] and f is continuous on the range of u=g(x) then

∫abfgxg'xdx=∫g(a)g(b)f(u)du

**3) Formula:**

∫xndx=xn+1n+1+C

**3) Given:**

∫02y21+y3dy

**4) Calculation:**

Consider, ∫02y21+y3dy

Since the function is continuous in the given interval, the integral exists.

By applying substitution rule for definite integrals

Let u=1+y3. Then du=3y2dy, so y2dy=du3

When y=0, u=1 and y=2, u=9

Thus,

∫02y21+y3dy=∫19u3du

By using fundamental theorem of calculus and power rule,

=u329219

=29u3219

Simplify,

=29[932-132]

=29(27-1)

=29(26)

=529

**Conclusion:**

Therefore, ∫02y21+y3dy=529