Calculadora de integrales definidas e impropias

La calculadora intentará evaluar la integral definida (es decir, con límites), incluida la impropia, con los pasos mostrados.

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Integrate with respect to:

Enter a lower limit:

If you need

$\displaystyle{\large{-\infty}}$ , type -inf.

Enter an upper limit:

If you need

$\displaystyle{\large{\infty}}$ , type inf.

Please write without any differentials such as $\displaystyle{\large{{d}{x}}}$ , $\displaystyle{\large{{d}{y}}}$ etc.

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Solution

Your input: calculate $\int_{0}^{2}\left( 3 x^{2} + x - 1 \right)dx$

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

$\left(x^{3} + \frac{x^{2}}{2} - x\right)|_{\left(x=2\right)}=8$

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

$\int_{0}^{2}\left( 3 x^{2} + x - 1 \right)dx=\left(x^{3} + \frac{x^{2}}{2} - x\right)|_{\left(x=2\right)}-\left(x^{3} + \frac{x^{2}}{2} - x\right)|_{\left(x=0\right)}=8$

Answer: $\int_{0}^{2}\left( 3 x^{2} + x - 1 \right)dx=8$

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