Animated Solution for Mathematics - Definite Integration: The area bounded by the curve $4y^2=x^2(4-x)(x-2)$ is equal to:

To find the area bounded by the curve $4y^2=x^2(4-x)(x-2)$, we must first determine the domain where the curve exists. Since $y^2$ must be non-negative, the expression $x^2(4-x)(x-2)$ must be $\ge 0$.

Because $x^2$ is always non-negative, the condition simplifies to $(4-x)(x-2)\ge 0$. Testing the intervals, we find that the curve exists only for $x\in [2,4]$.

The presence of $y^2$ in the equation indicates that the curve is symmetric about the $x$-axis. This allows us to calculate the area of the upper half and multiply by two.

The expression for the upper half is:

The total area $A$ is given by:

We simplify the quadratic expression inside the square root:

Substituting this into our integral, we get:

We apply the substitution $t=x-3$, which implies $dx=dt$. The limits of integration change from $[2,4]$ to $[-1,1]$:

We split the integral into two distinct parts:

The first term, ${\int }_{-1}^{1}t\sqrt{1-t^2}dt$, involves an odd function over a symmetric interval, which evaluates to $0$. We are left with the second term:

The integral ${\int }_0^1\sqrt{1-t^2}dt$ represents the area of a quarter-circle with radius $1$, which is $\frac{\pi }{4}$. Alternatively, using the standard integral formula:

Evaluating from $0$ to $1$:

The final area bounded by the curve is $\frac{3\pi }{2}$.