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calculates the exact area under the curve using Riemann sums of f(x)= 4-x2 on an interval [ 1,2]

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Answer to a math question calculates the exact area under the curve using Riemann sums of f(x)= 4-x2 on an interval [ 1,2]

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Para calcular el área exacta bajo la curva mediante sumas de Riemann de la función f(x) = 4 - x^2 en el intervalo [1,2], vamos a seguir los siguientes pasos:

Paso 1: Dividir el intervalo [1,2] en subintervalos iguales. Vamos a elegir n subintervalos para aproximar el área.

Paso 2: Calcular el ancho de cada subintervalo. Encuentra el valor de \Delta x dividiendo la longitud del intervalo (2 - 1) entre n, es decir, \Delta x = \frac{{2 - 1}}{n}.

Paso 3: Determinar los puntos de evaluación. Escoge un punto dentro de cada subintervalo para evaluar la función. Aquí utilizaremos el punto medio de cada subintervalo.

Paso 4: Calcular el área aproximada bajo la curva. Para cada subintervalo, calculamos el área de un rectángulo cuya altura es el valor de la función evaluada en el punto medio del subintervalo, y cuya base es el ancho del subintervalo. Luego, sumamos todas estas áreas de los rectángulos para obtener una aproximación del área bajo la curva.

Paso 5: Tomar el límite cuando n tiende a infinito. A medida que aumentamos el número de subintervalos, la aproximación del área se vuelve más precisa. Tomando el límite cuando n tiende a infinito, obtendremos el área exacta bajo la curva.

Ahora vamos a calcular el área exacta utilizando los métodos de sumas de Riemann:

Paso 1: Dividir el intervalo [1,2] en subintervalos iguales. Tomaremos n subintervalos.

Paso 2: Calcular el ancho de cada subintervalo. Tenemos \Delta x = \frac{{2 - 1}}{n} = \frac{1}{n}.

Paso 3: Determinar los puntos de evaluación. Utilizaremos el punto medio de cada subintervalo.

Paso 4: Calcular el área aproximada bajo la curva. Para cada subintervalo i, el punto de evaluación será x_i^* = 1 + \frac{\Delta x}{2} + i \cdot \Delta x, y el área del rectángulo correspondiente será A_i = f(x_i^*) \cdot \Delta x. Entonces, el área aproximada A será la suma de todas estas áreas:

A = \sum_{i=0}^{n-1} A_i = \sum_{i=0}^{n-1} f\left(1 + \frac{\Delta x}{2} + i \cdot \Delta x\right) \cdot \Delta x

El límite de esta suma cuando n tiende a infinito nos dará el área exacta bajo la curva.

Paso 5: Tomar el límite cuando n tiende a infinito. Es decir:

\lim_{{n \to \infty}} \sum_{i=0}^{n-1} f\left(1 + \frac{\Delta x}{2} + i \cdot \Delta x\right) \cdot \Delta x

Para calcular el límite de esta suma, podemos utilizar el teorema fundamental del cálculo o notar que la función f(x) = 4 - x^2 es continua en el intervalo [1,2] y, por lo tanto, integrable. Por lo tanto, el área exacta bajo la curva se puede calcular mediante la integral definida de la función en el intervalo [1,2]:

A = \int_{1}^{2} (4 - x^2) \, dx

Ahora podemos proceder a calcular la integral para obtener el área exacta:

\int_{1}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{1}^{2}

Evaluamos la integral en los límites de integración:

= \left[ 4(2) - \frac{(2)^3}{3} \right] - \left[ 4(1) - \frac{(1)^3}{3} \right]

= \left[ 8 - \frac{8}{3} \right] - \left[ 4 - \frac{1}{3} \right]

= \left[ \frac{24}{3} - \frac{8}{3} \right] - \left[ \frac{12}{3} - \frac{1}{3} \right]

= \frac{16}{3} - \frac{11}{3}

= \frac{5}{3}

Entonces, el área exacta bajo la curva f(x) = 4 - x^2 en el intervalo [1,2] es \frac{5}{3}.

\textbf{Respuesta:} El área exacta bajo la curva mediante sumas de Riemann de f(x) = 4 - x^2 en el intervalo [1,2] es \frac{5}{3}.

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