定積分與廣義積分計算器
逐步計算定積分與瑕積分
此計算器將嘗試計算定積分(即帶有上下限的積分),包含廣義積分,並顯示步驟。
Solution
Your input: calculate $$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x}=e^{- \frac{1}{x}}$$$ (for steps, see indefinite integral calculator)
Since there is infinity in the upper bound, this is improper integral of type 1.
To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (infinite).
$$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx=\lim_{x \to \infty}\left(e^{- \frac{1}{x}}\right)-\left(e^{- \frac{1}{x}}\right)|_{\left(x=6\right)}=1 - e^{- \frac{1}{6}}$$$
Answer: $$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx=1 - e^{- \frac{1}{6}}\approx 0.153518275109386$$$