# Definite and Improper Integral Calculator

The calculator will try to evaluate the definite (i.e. with bounds) integral, including improper, with steps shown.

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If you need -oo, type -inf.

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If you need oo, type inf.

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