Respuesta :
Answer:
[tex]\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558[/tex]
Step-by-step explanation:
We want to find the Riemann sum for [tex]\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx[/tex] with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:
[tex]\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)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex].
Step 1: Find [tex]\Delta{x}[/tex]
We have that [tex]a=0, b=\frac{3\pi }{4}, n=6[/tex]
Therefore, [tex]\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}[/tex]
Step 2: Divide the interval [tex]\left[0,\frac{3 \pi}{4}\right][/tex] into n = 6 sub-intervals of length [tex]\Delta{x}=\frac{\pi}{8}[/tex]
[tex]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[/tex]
Step 3: Evaluate the function at the left endpoints
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3[/tex]
[tex]f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386[/tex]
[tex]f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964[/tex]
[tex]f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527[/tex]
[tex]f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0[/tex]
[tex]f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527[/tex]
Step 4: Apply the Left Riemann Sum formula
[tex]\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315[/tex]
[tex]\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558[/tex]
Answer:
The Riemann sum with [tex]n=6[/tex] for [tex]f(x)=3 \mathrm cos x \,\,\,\,0\leq x\leq \frac{3\pi}{4}[/tex] is [tex]3.099558[/tex].
Step-by-step explanation:
The given integral function is,
[tex]\int\limits^\frac{3\pi }{4} _0 {3cos(x)} \, dx[/tex]
Use the left Riemann sum.
[tex]\int\limits^a_b {f(x)} \, dx \approx \Delta x (f(x_0)+f_1(x_1)+2f_2(x_2)+....+f(x_{n-2})+f(x_{n-1})[/tex]
Calculate the [tex]\Delta x[/tex].
[tex]\Delta x=\frac{b-a}{n}[/tex]
Put the values.
[tex]\Delta=\frac{\frac{3\pi }{4} -0}{6} \\=\frac{\pi }{8}[/tex]
The given interval is [tex](0.\frac{3\pi }{4} )[/tex]. Now divide into 6 sub interval of length [tex]\frac{\pi }{8}[/tex].
[tex]a=(0,\frac{\pi }{8}) ,(\frac{\pi }{8} ,\frac{\pi }{4}),(\frac{\pi }{4} ,\frac{3\pi }{8}),(\frac{3\pi }{8} ,\frac{\pi }{2}),(\frac{\pi }{2} ,\frac{5\pi }{8}),(\frac{5\pi }{8} ,\frac{3\pi }{4})=b[/tex]
Now, calculate the function at left end point.
[tex]f(x_0)=f(a)=f(0)=3\\f(x_1)=f(\frac{\pi }{8} )=3\sqrt{\frac{\sqrt{2} }{4}+\frac{1}{2} } =2.771638\\f(x_2)=f(\frac{\pi }{4} )=\frac{3\sqrt{2} }{2} =2.121320\\f(x_3)=f(\frac{3\pi }{8} )=3\sqrt{\frac{1 }{2}-\frac{\sqrt{2} }{4} } =1.148050\\f(x_4)=f(\frac{\pi }{2} )=0=0\\f(x_5)=f(\frac{5\pi }{8} )=-3\sqrt{\frac{1 }{2}-\frac{\sqrt{2} }{4} } =-1.148050[/tex]
Now, from Left Riemann Sum formula
[tex]\frac{\pi }{8} (3+2.771638+2.121320+1.148050+0-1.14805029=3.099557[/tex]
Therefore, Riemann sum with [tex]n=6[/tex] for [tex]f(x)=3 \mathrm cos x \,\,\,\,0\leq x\leq \frac{3\pi}{4}[/tex] is [tex]3.099558[/tex].
Learn more about Riemann here:
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