Knowee
Questions
Features
Study Tools

To find the area between the curves \(f(x) = x\) and \(f(x) = x^2\) in the interval \(0 \leq x \leq 1\), follow these steps: 1. **Set up the integral**: The area between the curves is given by the integral of the difference between the two functions over the given interval: \[ \int_0^1 (x - x^2) \, dx \] 2. **Evaluate the integral**: \[ \int_0^1 (x - x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx \] 3. **Find the antiderivatives**: \[ \int x \, dx = \frac{x^2}{2} + C \] \[ \int x^2 \, dx = \frac{x^3}{3} + C \] 4. **Apply the limits of integration**: \[ \left[ \frac{x^2}{2} \right]_0^1 - \left[ \frac{x^3}{3} \right]_0^1 \] 5. **Evaluate the definite integrals**: \[ \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \] \[ \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] 6. **Subtract the results**: \[ \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \] So, the area between the curves \(f(x) = x\) and \(f(x) = x^2\) in the interval \(0 \leq x \leq 1\) is \(\frac{1}{6}\). The correct answer is: - \(\frac{1}{6}\)

Question

To find the area between the curves f(x)=xf(x) = x and f(x)=x2f(x) = x^2 in the interval 0≤x≤10 \leq x \leq 1, follow these steps: 1. Set up the integral: The area between the curves is given by the integral of the difference between the two functions over the given interval: ∫01(x−x2) dx \int_0^1 (x - x^2) \, dx 2. Evaluate the integral: ∫01(x−x2) dx=∫01x dx−∫01x2 dx \int_0^1 (x - x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx 3. Find the antiderivatives: ∫x dx=x22+C \int x \, dx = \frac{x^2}{2} + C ∫x2 dx=x33+C \int x^2 \, dx = \frac{x^3}{3} + C 4. Apply the limits of integration: [x22]01−[x33]01 \left[ \frac{x^2}{2} \right]_0^1 - \left[ \frac{x^3}{3} \right]_0^1 5. Evaluate the definite integrals: [x22]01=122−022=12 \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} [x33]01=133−033=13 \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} 6. Subtract the results: 12−13=36−26=16 \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} So, the area between the curves f(x)=xf(x) = x and f(x)=x2f(x) = x^2 in the interval 0≤x≤10 \leq x \leq 1 is 16\frac{1}{6}. The correct answer is: - 16\frac{1}{6}

...expand
🧐 Not the exact question you are looking for?Go ask a question

Solution

Para encontrar el área entre las curvas f(x)=xf(x) = x y f(x)=x2f(x) = x^2 en el intervalo 0≤x≤10 \leq x \leq 1, sigue estos pasos:

  1. Configura la integral: El área entre las curvas se da por la integral de la diferencia entre las dos funciones sobre el intervalo dado: ∫01(x−x2) dx \int_0^1 (x - x^2) \, dx

  2. Evalúa la integral: ∫01(x−x2) dx=∫01x dx−∫01x2 dx \int_0^1 (x - x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx

  3. Encuentra las antiderivadas: ∫x dx=x22+C \int x \, dx = \frac{x^2}{2} + C ∫x2 dx=x33+C \int x^2 \, dx = \frac{x^3}{3} + C

  4. Aplica los límites de integración: [x22]01−[x33]01 \left[ \frac{x^2}{2} \right]_0^1 - \left[ \frac{x^3}{3} \right]_0^1

  5. Evalúa las integrales definidas: [x22]01=122−022=12 \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} [x33]01=133−033=13 \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}

  6. Resta los resultados: 12−13=36−26=16 \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}

Por lo tanto, el área entre las curvas f(x)=xf(x) = x y f(x)=x2f(x) = x^2 en el intervalo 0≤x≤10 \leq x \leq 1 es 16\frac{1}{6}. La respuesta correcta es: - 16\frac{1}{6}

This problem has been solved

Similar Questions

What is the area between two curves?

EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas.(a)    39 − x2 dx0(b)    10(x − 3) dx0SOLUTION (a) Since f(x) = 9 − x2 ≥ 0, we can interpret this integral as the area under the curve y = 9 − x2 from 0 to . But since y2 = , we get x2 + y2 = 9, which shows that the graph of f is a quarter-circle with radius in the top figure. Therefore,39 − x2 dx0 = 14𝜋(3)2 = .(b) The graph of y = x − 3 is the line with slope shown in the bottom figure. We compute the integral as the difference of the areas of the two triangles:10(x − 3) dx0 = A1 − A2 = − 4.5 = .

Estimate the area under the graph of f(x) = 2/x from x = 1 to x = 2 using four approximating rectangles and right endpoints. (Round your answer to four decimal places.)

Sketch the graph of the function 𝑥3:(a) Find the signed area between the curve, the 𝑥-axis, and the lines 𝑥=−2 and 𝑥=1.Signed area = (b) Find the geometric area between the curve, the 𝑥-axis, and the lines 𝑥=−2 and 𝑥=1.True area =

Use a suitable double integral to evaluate the area between the curve y = x2 and the curvey = x where x, y ≥ 0.

1/2

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.