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Harrizon [31]
3 years ago
6

f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans

wer to six decimal places.)
Mathematics
1 answer:
Kruka [31]3 years ago
5 0

Answer:

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

Step-by-step explanation:

We want to find the Riemann sum for \int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)

where \Delta{x}=\frac{b-a}{n}.

Step 1: Find \Delta{x}

We have that a=0, b=\frac{3\pi }{4}, n=6

Therefore, \Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}

Step 2: Divide the interval \left[0,\frac{3 \pi}{4}\right] into n = 6 sub-intervals of length \Delta{x}=\frac{\pi}{8}

a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b

Step 3: Evaluate the function at the left endpoints

f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3

f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386

f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964

f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527

f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0

f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527

Step 4: Apply the Left Riemann Sum formula

\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

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