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Find the work done in stretching a spring if force is given by F = 5x, from x = 0 to x = 2.

Mathematics

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Q8. Find the volume generated when y=x2y = x^2 is rotated about the x-axis from x=0x = 0 to x=1x = 1.

  1. Use the disk method formula for the volume of revolution about the x-axis

    V=π∫ab[f(x)]2 dxV = \pi \int_a^b [f(x)]^2 \, dx.
    Here, f(x)=x2f(x) = x^2, a=0a = 0, and b=1b = 1.

    V=π∫01(x2)2 dx=π∫01x4 dxV = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx
  2. Find the antiderivative of x4x^4.

    ∫x4 dx=x4+14+1=x55\int x^4 \, dx = \frac{x^{4+1}}{4+1} = \frac{x^5}{5}
  3. Evaluate the antiderivative at the limits of integration and multiply by π\pi.

    V=π[x55]01=π(155−055)V = \pi \left[\frac{x^5}{5}\right]_0^1 = \pi \left(\frac{1^5}{5} - \frac{0^5}{5}\right)
  4. Calculate the numerical value.

    V=π(15−0)=π5V = \pi \left(\frac{1}{5} - 0\right) = \frac{\pi}{5}

    The final answer is π5 cubic units\boxed{\frac{\pi}{5} \text{ cubic units}}.

    Q9. Find the work done in stretching a spring if force is given by F=5xF = 5x, from x=0x = 0 to x=2x = 2.

  5. The work done by a variable force is given by the definite integral of the force function with respect to displacement.

    W=∫abF(x) dxW = \int_a^b F(x) \, dx

    Here, F(x)=5xF(x) = 5x, a=0a = 0, and b=2b = 2.

    W=∫025x dxW = \int_0^2 5x \, dx
  6. Find the antiderivative of $5x$.

    ∫5x dx=5⋅x1+11+1=5⋅x22=52x2\int 5x \, dx = 5 \cdot \frac{x^{1+1}}{1+1} = 5 \cdot \frac{x^2}{2} = \frac{5}{2}x^2
  7. Evaluate the antiderivative at the limits of integration.

    W=[52x2]02=(52(2)2)−(52(0)2)W = \left[\frac{5}{2}x^2\right]_0^2 = \left(\frac{5}{2}(2)^2\right) - \left(\frac{5}{2}(0)^2\right)
  8. Calculate the numerical value.

    W=(52⋅4)−0=10W = \left(\frac{5}{2} \cdot 4\right) - 0 = 10

    The final answer is 10 units of work\boxed{10 \text{ units of work}}.

    Q10. Find the work done by a force F=3x2F = 3x^2 from x=1x = 1 to x=3x = 3.

  9. Set up the definite integral for the work done.

    W=∫abF(x) dxW = \int_a^b F(x) \, dx

    Here, F(x)=3x2F(x) = 3x^2, a=1a = 1, and b=3b = 3.

    W=∫133x2 dxW = \int_1^3 3x^2 \, dx
  10. Find the antiderivative of 3x23x^2.

    ∫3x2 dx=3⋅x2+12+1=3⋅x33=x3\int 3x^2 \, dx = 3 \cdot \frac{x^{2+1}}{2+1} = 3 \cdot \frac{x^3}{3} = x^3
  11. Evaluate the antiderivative at the limits of integration.

    W=[x3]13=(3)3−(1)3W = [x^3]_1^3 = (3)^3 - (1)^3
  12. Calculate the numerical value.

    W=27−1=26W = 27 - 1 = 26

    The final answer is 26 units of work\boxed{26 \text{ units of work}}.

    Q11. Find the average value of f(x)=x2f(x) = x^2 on [0,3][0,3].

  13. Use the formula for the average value of a function f(x)f(x) on an interval [a,b][a, b]

    favg=1b−a∫abf(x) dxf_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx.
    Here, f(x)=x2f(x) = x^2, a=0a = 0, and b=3b = 3.

    favg=13−0∫03x2 dx=13∫03x2 dxf_{avg} = \frac{1}{3-0} \int_0^3 x^2 \, dx = \frac{1}{3} \int_0^3 x^2 \, dx
  14. Find the antiderivative of x2x^2.

    ∫x2 dx=x2+12+1=x33\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}
  15. Evaluate the definite integral.

    ∫03x2 dx=[x33]03=(333)−(033)=273−0=9\int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \left(\frac{3^3}{3}\right) - \left(\frac{0^3}{3}\right) = \frac{27}{3} - 0 = 9
  16. Multiply by 1b−a\frac{1}{b-a}.

    favg=13⋅9=3f_{avg} = \frac{1}{3} \cdot 9 = 3

    The final answer is 3\boxed{3}.

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