정적분 및 가적분 계산기

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Solution

Your input: calculate $$$\int_{1}^{3}\left( \frac{\sqrt{1 - \frac{1}{x}}}{x^{2}} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{\frac{\sqrt{1 - \frac{1}{x}}}{x^{2}} d x}=\frac{2 \left(\frac{x - 1}{x}\right)^{\frac{3}{2}}}{3}$$$ (for steps, see indefinite integral calculator)

According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.

$$$\left(\frac{2 \left(\frac{x - 1}{x}\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=3\right)}=\frac{4 \sqrt{6}}{27}$$$

$$$\left(\frac{2 \left(\frac{x - 1}{x}\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=1\right)}=0$$$

$$$\int_{1}^{3}\left( \frac{\sqrt{1 - \frac{1}{x}}}{x^{2}} \right)dx=\left(\frac{2 \left(\frac{x - 1}{x}\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=3\right)}-\left(\frac{2 \left(\frac{x - 1}{x}\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=1\right)}=\frac{4 \sqrt{6}}{27}$$$

Answer: $$$\int_{1}^{3}\left( \frac{\sqrt{1 - \frac{1}{x}}}{x^{2}} \right)dx=\frac{4 \sqrt{6}}{27}\approx 0.362887369301212$$$

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정적분 및 가적분 계산기

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Solution