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Math  /  Calculus

QuestionUsing three rectangles and the midpoint rule, estimate the area above the xx-axis and under the function f(x)=1−(cos⁡πx3)6f(x)=1-\left(\cos \frac{\pi x}{3}\right)^{6} over the interval [0,3][0,3]. Leave your answer as an exact value. Note that f(x)≥0f(x) \geq 0 for all xx.

Studdy Solution

STEP 1

1. We are using the midpoint rule to estimate the area under the curve.
2. The interval [0,3][0, 3] is divided into three equal subintervals.
3. The function f(x)=1−(cos⁡πx3)6 f(x) = 1 - \left(\cos \frac{\pi x}{3}\right)^6 is non-negative over the interval.

STEP 2

1. Divide the interval [0,3][0, 3] into three equal subintervals.
2. Determine the midpoints of each subinterval.
3. Evaluate the function f(x) f(x) at each midpoint.
4. Calculate the area of each rectangle using the midpoint rule.
5. Sum the areas of the rectangles to estimate the total area.

STEP 3

Divide the interval [0,3][0, 3] into three equal subintervals. Since the interval length is 3, each subinterval will have a length of:
Δx=3−03=1 \Delta x = \frac{3 - 0}{3} = 1
The subintervals are [0,1][0, 1], [1,2][1, 2], and [2,3][2, 3].

STEP 4

Determine the midpoints of each subinterval:
- For [0,1][0, 1], the midpoint is x1=0+12=0.5 x_1 = \frac{0 + 1}{2} = 0.5 . - For [1,2][1, 2], the midpoint is x2=1+22=1.5 x_2 = \frac{1 + 2}{2} = 1.5 . - For [2,3][2, 3], the midpoint is x3=2+32=2.5 x_3 = \frac{2 + 3}{2} = 2.5 .

STEP 5

Evaluate the function f(x) f(x) at each midpoint:
- f(x1)=1−(cos⁡π×0.53)6=1−(cos⁡π6)6=1−(32)6 f(x_1) = 1 - \left(\cos \frac{\pi \times 0.5}{3}\right)^6 = 1 - \left(\cos \frac{\pi}{6}\right)^6 = 1 - \left(\frac{\sqrt{3}}{2}\right)^6 - f(x2)=1−(cos⁡π×1.53)6=1−(cos⁡π2)6=1−06=1 f(x_2) = 1 - \left(\cos \frac{\pi \times 1.5}{3}\right)^6 = 1 - \left(\cos \frac{\pi}{2}\right)^6 = 1 - 0^6 = 1 - f(x3)=1−(cos⁡π×2.53)6=1−(cos⁡5π6)6=1−(−32)6 f(x_3) = 1 - \left(\cos \frac{\pi \times 2.5}{3}\right)^6 = 1 - \left(\cos \frac{5\pi}{6}\right)^6 = 1 - \left(-\frac{\sqrt{3}}{2}\right)^6

STEP 6

Calculate the area of each rectangle using the midpoint rule. The area of each rectangle is given by f(xi)×Δx f(x_i) \times \Delta x :
- Area of rectangle 1: f(x1)×1=(1−(32)6)×1 f(x_1) \times 1 = \left(1 - \left(\frac{\sqrt{3}}{2}\right)^6\right) \times 1 - Area of rectangle 2: f(x2)×1=1×1=1 f(x_2) \times 1 = 1 \times 1 = 1 - Area of rectangle 3: f(x3)×1=(1−(−32)6)×1 f(x_3) \times 1 = \left(1 - \left(-\frac{\sqrt{3}}{2}\right)^6\right) \times 1

STEP 7

Sum the areas of the rectangles to estimate the total area:
Total Area=(1−(32)6)+1+(1−(−32)6)\text{Total Area} = \left(1 - \left(\frac{\sqrt{3}}{2}\right)^6\right) + 1 + \left(1 - \left(-\frac{\sqrt{3}}{2}\right)^6\right)
Since (32)6=(−32)6\left(\frac{\sqrt{3}}{2}\right)^6 = \left(-\frac{\sqrt{3}}{2}\right)^6, we can simplify:
Total Area=2+1−2(32)6\text{Total Area} = 2 + 1 - 2\left(\frac{\sqrt{3}}{2}\right)^6
The estimated area under the curve using the midpoint rule is:
3−2(32)6\boxed{3 - 2\left(\frac{\sqrt{3}}{2}\right)^6}

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