#### To determine

**To find:**

The area enclosed by the loop

#### Answer

2453

#### Explanation

**1) Concept:**

The substitution rule: If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here,g(x) is substituted as u and then differentiation g’(x)dx =du.

**2) Given:**

y2=x2(x+3)

**3) Formula:**

∫xn dx=xn+1n+1+C

4) **Calculation:**

From the graph, observe that the loop extends from x=-3 to x=0

And by symmetry, the area enclosed by the loop is twice the area under the top half of the curve on this interval.

The equation of top half is y=-xx+3.

Therefore, the area enclosed by the loop is

2∫-30-xx+3dx

Use substitution method.

Substitute x+3=u.

Therefore, differentiation is dx=du.

Changing the limits of integration,

At x=-3, u=-3+3=0 and

At x=0, u=0+3=3

Therefore, the integral from 0 to 3 becomes

2∫-30-xx+3dx=-2∫03(u-3)udu

Simplify.

=-2∫03(u32-3u12)du

Simplify.

=-225u52-2u3203

Substitute limits.

=-225352-2332-0

Simplify.

=2453

Therefore, the area enclosed by the loop is 2453

**Conclusion:**

The area enclosed by the loop is 2453