Calcolatore di integrali definiti e impropri

Il calcolatore cercherà di valutare l'integrale definito (cioè con estremi), inclusi quelli impropri, mostrando i passaggi.

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Solution

Your input: calculate $$$\int_{\frac{1}{3}}^{\frac{1}{2}}\left( -6 + \frac{1}{t^{3}} \right)dt$$$

First, calculate the corresponding indefinite integral: $$$\int{\left(-6 + \frac{1}{t^{3}}\right)d t}=- 6 t - \frac{1}{2 t^{2}}$$$ (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.

$$$\left(- 6 t - \frac{1}{2 t^{2}}\right)|_{\left(t=\frac{1}{2}\right)}=-5$$$

$$$\left(- 6 t - \frac{1}{2 t^{2}}\right)|_{\left(t=\frac{1}{3}\right)}=- \frac{13}{2}$$$

$$$\int_{\frac{1}{3}}^{\frac{1}{2}}\left( -6 + \frac{1}{t^{3}} \right)dt=\left(- 6 t - \frac{1}{2 t^{2}}\right)|_{\left(t=\frac{1}{2}\right)}-\left(- 6 t - \frac{1}{2 t^{2}}\right)|_{\left(t=\frac{1}{3}\right)}=\frac{3}{2}$$$

Answer: $$$\int_{\frac{1}{3}}^{\frac{1}{2}}\left( -6 + \frac{1}{t^{3}} \right)dt=\frac{3}{2}=1.5$$$

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Calcolatore di integrali definiti e impropri

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Solution