\( \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 - x ^ { 2 } } d x \)

To tackle the integral \( \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 - x ^ { 2 } } d x \), it's a good idea to first consider whether the limits make sense. Notice that for \( x \) from 0 to 2, \( \sqrt{1 - x^2} \) becomes imaginary once \( x > 1 \). This means that this integral is only defined on the interval where \( 0 \leq x \leq 1 \). Now, let's adjust the integral: \[ \int _ { 0 } ^ { 1 } x ^ { 3 } \sqrt { 1 - x ^ { 2 } } d x \] To solve this integral, we can use the substitution \( u = 1 - x^2 \), which leads to \( du = -2x\,dx \) or \( dx = -\frac{1}{2x} du \). When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 0 \). The integral now transforms: \[ -\frac{1}{2} \int_1^0 (1 - u)^{3/2} \sqrt{u} \cdot \frac{1}{\sqrt{1-u}} \, du \to \frac{1}{2} \int_0^1 (1 - u)^{3/2} \sqrt{u} \, du \] This integral can be computed using beta functions or direct integration techniques. After some calculations, you'll find the integral evaluates to a specific value (you may apply numerical methods or further transformations to get an exact solution depending on the method you prefer). The answer will give you a clearer representation of your original integrand’s area under the curve.