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 $$$\int_{1}^{e}\left( x^{3} \ln{\left(x \right)} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{x^{3} \ln{\left(x \right)} d x}=\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}$$$ (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.

$$$\left(\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}\right)|_{\left(x=e\right)}=\frac{3 e^{4}}{16}$$$

$$$\left(\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}\right)|_{\left(x=1\right)}=- \frac{1}{16}$$$

$$$\int_{1}^{e}\left( x^{3} \ln{\left(x \right)} \right)dx=\left(\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}\right)|_{\left(x=e\right)}-\left(\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}\right)|_{\left(x=1\right)}=\frac{1}{16} + \frac{3 e^{4}}{16}$$$

Answer: $$$\int_{1}^{e}\left( x^{3} \ln{\left(x \right)} \right)dx=\frac{1}{16} + \frac{3 e^{4}}{16}\approx 10.2996531312145$$$

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Solution