Question

Evaluate \[ \int_{0}^{2}x^{3}(4-x^{2})^{\frac52}\,dx \]

• A. \(\dfrac{256}{63}\)
• B. \(\dfrac{2048}{63}\)
• C. \(\dfrac{512}{63}\)
• D. \(\dfrac{1024}{63}\)

Hint:

For integrals involving \[ x^{m}(a-x^{2})^{n}, \] use the substitution \[ \boxed{u=a-x^{2}.} \] This converts the integral into a polynomial in \(u\), making it easy to evaluate.

Answer 1

The Correct Option is D

Step 1: Use substitution. Let \[ u=4-x^{2}. \] Then, \[ du=-2x\,dx, \] or \[ x\,dx=-\frac12\,du. \] Also, \[ x^{2}=4-u. \] Hence, \[ x^{3}dx = x^{2}(x\,dx) = -\frac12(4-u)\,du. \] The limits become \[ x=0\Rightarrow u=4, \] \[ x=2\Rightarrow u=0. \] Therefore, \[ I = \frac12 \int_{0}^{4}(4-u)u^{5/2}\,du. \]

Step 2: Expand and integrate. \[ I = \frac12 \left[ 4\int_{0}^{4}u^{5/2}\,du - \int_{0}^{4}u^{7/2}\,du \right]. \] Using \[ \int u^{5/2}du=\frac{2}{7}u^{7/2}, \] and \[ \int u^{7/2}du=\frac{2}{9}u^{9/2}, \] we obtain \[ I = \frac12 \left[ \frac87u^{7/2} - \frac29u^{9/2} \right]_{0}^{4}. \]

Step 3: Substitute the limits. Since \[ 4^{7/2}=128, \qquad 4^{9/2}=512, \] we get \[ I = \frac12 \left( \frac{1024}{7} - \frac{1024}{9} \right). \] \[ = \frac12 \cdot 1024 \left( \frac{2}{63} \right) = \frac{1024}{63}. \] Hence, \[ \boxed{ \int_{0}^{2}x^{3}(4-x^{2})^{5/2}\,dx = \frac{1024}{63}. } \] Therefore, the correct option is \(\boxed{(D)}\).