Question - Definite Integral of a Function
Solution:
To solve this definite integral, we need to integrate each term separately within the bounds from -1 to 0. Let's integrate the function $$ f(x) = 2x - e^x $$.The integral of $$ 2x $$ is $$ x^2 $$, and the integral of $$ -e^x $$ is $$ -e^x $$, since the derivative of $$ e^x $$ with respect to $$ x $$ is $$ e^x $$.Let's integrate and then apply the bounds:$$\int_{-1}^{0} (2x - e^x) dx = \left[ x^2 - e^x \right]_{-1}^{0}$$Now, applying the Fundamental Theorem of Calculus, we evaluate this antiderivative at the upper bound x=0 and subtract the value of the antiderivative at the lower bound x=-1:At x=0:$$x^2 - e^x = 0^2 - e^0 = 0 - 1 = -1$$At x=-1:$$x^2 - e^x = (-1)^2 - e^{-1} = 1 - \frac{1}{e}$$Now subtract the value at x=-1 from the value at x=0:$$(-1)-(1 - \frac{1}{e}) = -1 - 1 + \frac{1}{e} = -2 + \frac{1}{e}$$Hence the value of the definite integral is:$$\int_{-1}^{0} (2x - e^x) dx = -2 + \frac{1}{e}$$
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