Example Question - fundamental theorem of calculus
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Solving Definite Integral Using Fundamental Theorem of Calculus
Para resolver esta integral definida, aplicaremos el teorema fundamental del cálculo. La integral de x con respecto a x es \( \frac{x^2}{2} \), así que aplicamos los límites de integración: \( \int_{0}^{a} xdx = \left[\frac{x^2}{2}\right]_{0}^{a} \) Evaluamos esta expresión en los límites superior e inferior: \( \left[\frac{x^2}{2}\right]_{0}^{a} = \frac{a^2}{2} - \frac{0^2}{2} \) \( = \frac{a^2}{2} - 0 \) \( = \frac{a^2}{2} \) Por lo tanto, la respuesta correcta es la opción \( \text{d) } \frac{a^2}{2} \).
Definite Integral of a Function
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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