$$ \int _ { 0 } ^ { 2 } 1 \sqrt { 1 + x } d x \quad n = 4 $$

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Calculate or simplify the expression $$ \int_{0}^{2} \sqrt{1+x} dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{2} \sqrt{1+x} dx$$
- step1: Evaluate the integral:
  $$\int \sqrt{1+x} dx$$
- step2: Rewrite the expression:
  $$\int \left(1+xight)^{\frac{1}{2}} dx$$
- step3: Use the substitution $$dx=1 dt$$ to transform the integral$$:$$
  $$\int \left(1+xight)^{\frac{1}{2}}	imes 1 dt$$
- step4: Simplify:
  $$\int \left(1+xight)^{\frac{1}{2}} dt$$
- step5: Use the substitution $$t=1+x$$ to transform the integral$$:$$
  $$\int t^{\frac{1}{2}} dt$$
- step6: Evaluate the integral:
  $$\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$
- step7: Add the numbers:
  $$\frac{t^{\frac{3}{2}}}{\frac{1}{2}+1}$$
- step8: Add the numbers:
  $$\frac{t^{\frac{3}{2}}}{\frac{3}{2}}$$
- step9: Multiply by the reciprocal:
  $$t^{\frac{3}{2}}	imes \frac{2}{3}$$
- step10: Reorder the terms:
  $$\frac{2}{3}t^{\frac{3}{2}}$$
- step11: Substitute back:
  $$\frac{2}{3}\left(1+xight)^{\frac{3}{2}}$$
- step12: Return the limits:
  $$\left(\frac{2}{3}\left(1+xight)^{\frac{3}{2}}ight)\bigg |_{0}^{2}$$
- step13: Calculate the value:
  $$2	imes 3^{\frac{1}{2}}-\frac{2}{3}$$
- step14: Transform the expression:
  $$2\sqrt{3}-\frac{2}{3}$$
- step15: Calculate:
  $$\frac{6\sqrt{3}-2}{3}$$
La integral de $$ \int_{0}^{2} \sqrt{1+x} dx $$ es aproximadamente 2.797435.
La integral de $$ \sqrt{1+x} $$ desde 0 hasta 2 es aproximadamente 2.797.

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