题目
类似地，求极限lim_(ntoinfty)nint_(0)^1x^nsqrt(1+x^2)dx.
类似地，
求极限$\lim_{n\to\infty}n\int_{0}^{1}x^{n}\sqrt{1+x^{2}}dx$.
题目解答
答案
使用分部积分法，设 $u = \sqrt{1 + x^2}$，$dv = x^n \, dx$，则 $du = \frac{x}{\sqrt{1 + x^2}} \, dx$，$v = \frac{x^{n+1}}{n+1}$。分部积分得：
\[
\int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \frac{\sqrt{2}}{n+1} - \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx.
\]
乘以 $n$ 并取极限，利用 $\lim_{n \to \infty} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx = 0$，得：
\[
\lim_{n \to \infty} n \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \sqrt{2}.
\]
或者，由连续函数 $f(x)$ 的性质：
\[
\lim_{n \to \infty} n \int_{0}^{1} x^n f(x) \, dx = f(1),
\]
取 $f(x) = \sqrt{1 + x^2}$，得 $f(1) = \sqrt{2}$。
**答案：** $\boxed{\sqrt{2}}$
解析
步骤 1：分部积分法
设 $u = \sqrt{1 + x^2}$，$dv = x^n \, dx$，则 $du = \frac{x}{\sqrt{1 + x^2}} \, dx$，$v = \frac{x^{n+1}}{n+1}$。根据分部积分法，我们有：
\[ \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \left[ \frac{x^{n+1}}{n+1} \sqrt{1 + x^2} \right]_{0}^{1} - \int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{x}{\sqrt{1 + x^2}} \, dx. \]
步骤 2：计算边界值
计算边界值，我们得到：
\[ \left[ \frac{x^{n+1}}{n+1} \sqrt{1 + x^2} \right]_{0}^{1} = \frac{1}{n+1} \sqrt{2} - 0 = \frac{\sqrt{2}}{n+1}. \]
步骤 3：简化积分
简化积分，我们得到：
\[ \int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{x}{\sqrt{1 + x^2}} \, dx = \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx. \]
步骤 4：乘以 $n$ 并取极限
乘以 $n$ 并取极限，我们得到：
\[ \lim_{n \to \infty} n \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \lim_{n \to \infty} n \left( \frac{\sqrt{2}}{n+1} - \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx \right). \]
步骤 5：利用极限性质
利用 $\lim_{n \to \infty} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx = 0$，我们得到：
\[ \lim_{n \to \infty} n \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \lim_{n \to \infty} \frac{n \sqrt{2}}{n+1} = \sqrt{2}. \]
设 $u = \sqrt{1 + x^2}$，$dv = x^n \, dx$，则 $du = \frac{x}{\sqrt{1 + x^2}} \, dx$，$v = \frac{x^{n+1}}{n+1}$。根据分部积分法，我们有：
\[ \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \left[ \frac{x^{n+1}}{n+1} \sqrt{1 + x^2} \right]_{0}^{1} - \int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{x}{\sqrt{1 + x^2}} \, dx. \]
步骤 2：计算边界值
计算边界值，我们得到：
\[ \left[ \frac{x^{n+1}}{n+1} \sqrt{1 + x^2} \right]_{0}^{1} = \frac{1}{n+1} \sqrt{2} - 0 = \frac{\sqrt{2}}{n+1}. \]
步骤 3：简化积分
简化积分，我们得到：
\[ \int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{x}{\sqrt{1 + x^2}} \, dx = \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx. \]
步骤 4：乘以 $n$ 并取极限
乘以 $n$ 并取极限，我们得到：
\[ \lim_{n \to \infty} n \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \lim_{n \to \infty} n \left( \frac{\sqrt{2}}{n+1} - \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx \right). \]
步骤 5：利用极限性质
利用 $\lim_{n \to \infty} \int_{0}^{1} \frac{x^{n+2}}{\sqrt{1 + x^2}} \, dx = 0$，我们得到：
\[ \lim_{n \to \infty} n \int_{0}^{1} x^n \sqrt{1 + x^2} \, dx = \lim_{n \to \infty} \frac{n \sqrt{2}}{n+1} = \sqrt{2}. \]