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Approximate the definite integral by right Riemann sum with the indicated partitions.\newliney=cosxy=\cos x\newline[0,π6,π4,π2,3π4,π][0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi]

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Q. Approximate the definite integral by right Riemann sum with the indicated partitions.\newliney=cosxy=\cos x\newline[0,π6,π4,π2,3π4,π][0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi]
  1. Problem and Riemann sum setup: Understand the problem and set up the Riemann sum. We are asked to approximate the definite integral of the function y=cos(x)y = \cos(x) over the interval [0,π][0, \pi] using a right Riemann sum. This means we will use the value of the function at the right endpoint of each subinterval to calculate the sum. The partitions given are [0,π/6,π/4,π/2,3π/4,π][0, \pi/6, \pi/4, \pi/2, 3\pi/4, \pi], which divide the interval [0,π][0, \pi] into 55 subintervals. We will calculate the width of each subinterval and the value of the function at the right endpoint of each subinterval.
  2. Calculate subinterval widths: Calculate the width of each subinterval.\newlineThe widths of the subintervals are as follows:\newline- From π/6\pi/6 to π/4\pi/4: π/4π/6=π/12\pi/4 - \pi/6 = \pi/12\newline- From π/4\pi/4 to π/2\pi/2: π/2π/4=π/4\pi/2 - \pi/4 = \pi/4\newline- From π/2\pi/2 to 3π/43\pi/4: 3π/4π/2=π/43\pi/4 - \pi/2 = \pi/4\newline- From 3π/43\pi/4 to π/4\pi/400: π/4\pi/411\newlineNote that the first subinterval (from π/4\pi/422 to π/6\pi/6) is not used in the right Riemann sum because we are using the right endpoints, and π/4\pi/422 is the left endpoint of the first subinterval.
  3. Evaluate function at right endpoints: Evaluate the function at the right endpoints of each subinterval.\newlineWe will evaluate cos(x)\cos(x) at the right endpoints of each subinterval:\newline- cos(π4)=22\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\newline- cos(π2)=0\cos(\frac{\pi}{2}) = 0\newline- cos(3π4)=22\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}\newline- cos(π)=1\cos(\pi) = -1
  4. Multiply function values by subinterval widths: Multiply each function value by the width of its subinterval.\newlineNow we multiply the value of the function at each right endpoint by the width of the corresponding subinterval:\newline- cos(π4)×(π4π6)=(22)×(π12)\cos(\frac{\pi}{4}) \times (\frac{\pi}{4} - \frac{\pi}{6}) = (\frac{\sqrt{2}}{2}) \times (\frac{\pi}{12})\newline- cos(π2)×(π2π4)=0×(π4)\cos(\frac{\pi}{2}) \times (\frac{\pi}{2} - \frac{\pi}{4}) = 0 \times (\frac{\pi}{4})\newline- cos(3π4)×(3π4π2)=(22)×(π4)\cos(\frac{3\pi}{4}) \times (\frac{3\pi}{4} - \frac{\pi}{2}) = (-\frac{\sqrt{2}}{2}) \times (\frac{\pi}{4})\newline- cos(π)×(π3π4)=1×(π4)\cos(\pi) \times (\pi - \frac{3\pi}{4}) = -1 \times (\frac{\pi}{4})
  5. Calculate right Riemann sum: Add up the products to get the right Riemann sum. The right Riemann sum is the sum of all the products calculated in the previous step: R=(2/2)(π/12)+0(π/4)+(2/2)(π/4)+(1)(π/4)R = (\sqrt{2}/2) \cdot (\pi/12) + 0 \cdot (\pi/4) + (-\sqrt{2}/2) \cdot (\pi/4) + (-1) \cdot (\pi/4) R=(2/2)(π/12)(2/2)(π/4)(π/4)R = (\sqrt{2}/2) \cdot (\pi/12) - (\sqrt{2}/2) \cdot (\pi/4) - (\pi/4)
  6. Simplify Riemann sum: Simplify the Riemann sum.\newlineTo simplify the Riemann sum, we combine like terms and perform the arithmetic:\newlineR=(2/2)×(π/12)(2/2)×(3π/12)(π/4)R = (\sqrt{2}/2) \times (\pi/12) - (\sqrt{2}/2) \times (3\pi/12) - (\pi/4)\newlineR=(2/2)×(π/123π/12)(π/4)R = (\sqrt{2}/2) \times (\pi/12 - 3\pi/12) - (\pi/4)\newlineR=(2/2)×(2π/12)(π/4)R = (\sqrt{2}/2) \times (-2\pi/12) - (\pi/4)\newlineR=(2/6)×(π)(π/4)R = (-\sqrt{2}/6) \times (\pi) - (\pi/4)\newlineR=2π/6π/4R = -\sqrt{2}\pi/6 - \pi/4

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