(2) $$ \int_{0}^{2} \frac{x}{\sqrt{16-x^{2}}} d x $$

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Calculate or simplify the expression $$ \int_{0}^{2} \frac{x}{\sqrt{16-x^{2}}} d x $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{2} \frac{x}{\sqrt{16-x^{2}}} dx$$
- step1: Evaluate the integral:
  $$\int \frac{x}{\sqrt{16-x^{2}}} dx$$
- step2: Rewrite the expression:
  $$\int \frac{x}{\left(16-x^{2}ight)^{\frac{1}{2}}} dx$$
- step3: Use the substitution $$dx=-\frac{1}{2x} dt$$ to transform the integral$$:$$
  $$\int \frac{x}{\left(16-x^{2}ight)^{\frac{1}{2}}}	imes \left(-\frac{1}{2x}ight) dt$$
- step4: Simplify:
  $$\int -\frac{1}{2\left(16-x^{2}ight)^{\frac{1}{2}}} dt$$
- step5: Use the substitution $$t=-x^{2}$$ to transform the integral$$:$$
  $$\int \frac{-1}{2\left(16+tight)^{\frac{1}{2}}} dt$$
- step6: Rewrite the expression:
  $$\int -\frac{1}{2}	imes \frac{1}{\left(16+tight)^{\frac{1}{2}}} dt$$
- step7: Use properties of integrals:
  $$-\frac{1}{2}	imes \int \frac{1}{\left(16+tight)^{\frac{1}{2}}} dt$$
- step8: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$-\frac{1}{2}	imes \int \frac{1}{\left(16+tight)^{\frac{1}{2}}}	imes 1 dv$$
- step9: Simplify:
  $$-\frac{1}{2}	imes \int \frac{1}{\left(16+tight)^{\frac{1}{2}}} dv$$
- step10: Use the substitution $$v=16+t$$ to transform the integral$$:$$
  $$-\frac{1}{2}	imes \int \frac{1}{v^{\frac{1}{2}}} dv$$
- step11: Evaluate the integral:
  $$-\frac{1}{2}	imes \frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$
- step12: Simplify:
  $$-\frac{1}{2}	imes \frac{v^{\frac{1}{2}}}{\frac{1}{2}}$$
- step13: Multiply the terms:
  $$-v^{\frac{1}{2}}$$
- step14: Substitute back:
  $$-\left(16+tight)^{\frac{1}{2}}$$
- step15: Substitute back:
  $$-\left(16-x^{2}ight)^{\frac{1}{2}}$$
- step16: Return the limits:
  $$\left(-\left(16-x^{2}ight)^{\frac{1}{2}}ight)\bigg |_{0}^{2}$$
- step17: Calculate the value:
  $$-12^{\frac{1}{2}}+4$$
- step18: Calculate:
  $$-2\sqrt{3}+4$$
La integral de $$ \frac{x}{\sqrt{16-x^{2}}} $$ desde 0 hasta 2 es igual a $$ -2\sqrt{3}+4 $$.
La integral de $$ \frac{x}{\sqrt{16-x^{2}}} $$ desde 0 hasta 2 es $$ -2\sqrt{3}+4 $$.

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