Math Problem Statement

Find the mean, median and mode of f(x) = sin x[0,pi/2]

Solution

To find the mean, median, and mode of the function f(x)=sin(x)f(x) = \sin(x) on the interval [0,π2]\left[ 0, \frac{\pi}{2} \right], we'll follow these steps:

1. Mean

The mean of a function over an interval is given by:

Mean=1baabf(x)dx\text{Mean} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx

For f(x)=sin(x)f(x) = \sin(x) on the interval [0,π2]\left[ 0, \frac{\pi}{2} \right], we have:

Mean=1π200π2sin(x)dx=2π0π2sin(x)dx\text{Mean} = \frac{1}{\frac{\pi}{2} - 0} \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx

We know that the integral of sin(x)\sin(x) is cos(x)-\cos(x), so:

0π2sin(x)dx=[cos(x)]0π2=cos(π2)+cos(0)=0+1=1\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \left[ -\cos(x) \right]_{0}^{\frac{\pi}{2}} = -\cos\left( \frac{\pi}{2} \right) + \cos(0) = 0 + 1 = 1

Thus, the mean is:

Mean=2π×1=2π\text{Mean} = \frac{2}{\pi} \times 1 = \frac{2}{\pi}

2. Median

The median is the value mm such that half of the total area under the curve lies to the left of mm. Mathematically, we find mm such that:

0msin(x)dx=120π2sin(x)dx\int_0^m \sin(x) \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(x) \, dx

We already computed that the total area under the curve is 1, so we want:

0msin(x)dx=12\int_0^m \sin(x) \, dx = \frac{1}{2}

Solving this:

cos(m)+cos(0)=12    cos(m)+1=12    cos(m)=12-\cos(m) + \cos(0) = \frac{1}{2} \implies -\cos(m) + 1 = \frac{1}{2} \implies \cos(m) = \frac{1}{2}

So:

m=cos1(12)=π3m = \cos^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{3}

Thus, the median is π3\frac{\pi}{3}.

3. Mode

The mode is the value of xx that maximizes the function. Since sin(x)\sin(x) is increasing on [0,π2]\left[ 0, \frac{\pi}{2} \right], the maximum value occurs at x=π2x = \frac{\pi}{2}, so the mode is:

Mode=π2\text{Mode} = \frac{\pi}{2}

Summary:

  • Mean: 2π\frac{2}{\pi}
  • Median: π3\frac{\pi}{3}
  • Mode: π2\frac{\pi}{2}

Would you like further clarification or more details?

Here are 5 related questions for you to consider:

  1. How would the mean change if the interval was [0,π]\left[0, \pi \right]?
  2. What is the effect of modifying the function to f(x)=cos(x)f(x) = \cos(x) on the same interval?
  3. Can you find the variance of f(x)=sin(x)f(x) = \sin(x) on the interval [0,π2]\left[ 0, \frac{\pi}{2} \right]?
  4. What is the geometric interpretation of the median in this context?
  5. How does the shape of the sin(x)\sin(x) function affect its mode?

Tip: When calculating means over an interval, always remember to divide by the length of the interval after integrating.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Definite Integrals
Function Analysis

Formulas

Mean of a function: (1/(b-a)) * ∫[a,b] f(x) dx
Integral of sin(x): -cos(x) + C
Cosine inverse function: cos⁻¹(x)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Calculus I/II)