Find Antiderivative: We need to find the definite integral of the function (2x-1) from -1 to 0. The first step is to find the antiderivative of the function. The antiderivative of 2x is x^2, and the antiderivative of -1 is -x. Therefore, the antiderivative of (2x-1) is x^2 - x.Evaluate at Limits: Now we will evaluate the antiderivative at the upper and lower limits of the integral and subtract the two values. The antiderivative evaluated at the upper limit (x=0) is (0)^2 - (0) = 0. The antiderivative evaluated at the lower limit (x=-1) is (-1)^2 - (-1) = 1 + 1 = 2.Subtract Values: Subtract the value of the antiderivative at the lower limit from the value at the upper limit to find the definite integral. The definite integral from -1 to 0 of (2x-1) is 0 - 2 = -2.

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$\int_{-1}^{0}(2 x-1) d x$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{-1}^{0}(2 x-1) d x

Find Antiderivative: We need to find the definite integral of the function $(2x-1)$ from $-1$ to $0$ . The first step is to find the antiderivative of the function. $\newline$ The antiderivative of $2x$ is $x^2$ , and the antiderivative of $-1$ is $-x$ . Therefore, the antiderivative of $(2x-1)$ is $x^2 - x$ .
Evaluate at Limits: Now we will evaluate the antiderivative at the upper and lower limits of the integral and subtract the two values. $\newline$ The antiderivative evaluated at the upper limit $x=0$ is $(0)^2 - (0) = 0$ . $\newline$ The antiderivative evaluated at the lower limit $x=-1$ is $(-1)^2 - (-1) = 1 + 1 = 2$ .
Subtract Values: Subtract the value of the antiderivative at the lower limit from the value at the upper limit to find the definite integral. $\newline$ The definite integral from $-1$ to $0$ of $(2x-1)$ is $0 - 2 = -2$ .