2. evaluate ∫₀² 2x√(x² + 1)dx

Explanation:

Step1: Use substitution

Let $u = x^{2}+1$, then $du=2xdx$. When $x = 0$, $u=0^{2}+1 = 1$; when $x = 2$, $u=2^{2}+1=5$. The integral $\int_{0}^{2}2x\sqrt{x^{2}+1}dx$ becomes $\int_{1}^{5}\sqrt{u}du$.

Step2: Rewrite the integrand

$\sqrt{u}=u^{\frac{1}{2}}$, so the integral is $\int_{1}^{5}u^{\frac{1}{2}}du$.

Step3: Apply the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $n=\frac{1}{2}$, we have $\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{2}{3}u^{\frac{3}{2}}+C$.

Step4: Evaluate the definite integral

$\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{5}=\frac{2}{3}\times5^{\frac{3}{2}}-\frac{2}{3}\times1^{\frac{3}{2}}=\frac{2}{3}(5\sqrt{5}-1)$.

Answer:

$\frac{2}{3}(5\sqrt{5}-1)$