The solution for l4 is mathematically given as
L_{4}=0.5431
<h3>What is the solution for l4?</h3>
![&\text { Given } f(x)=a \cos ^{2}(x) \quad\left[\frac{\pi}{8}, \frac{\pi}{2}\right] \text { \& } n=4\\\\&\Delta a=\frac{b-a}{n}=\frac{\pi / 2-\pi / 8}{9}=\frac{3 \pi}{32}\\\\&x_{0}=\pi / 8, x_{1}=\frac{\pi}{8}+\frac{3 \pi}{32}=\frac{7 \pi}{32}\\\\&x_{2}=\frac{5 \pi}{16}, \quad x_{3}=\frac{13 \pi}{32}, x_{4}=\pi / 2\\\\&f\left(x_{0}\right)=f(1 / 8)=0.8535\\](https://tex.z-dn.net/?f=%26%5Ctext%20%7B%20Given%20%7D%20f%28x%29%3Da%20%5Ccos%20%5E%7B2%7D%28x%29%20%5Cquad%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%20%5Ctext%20%7B%20%5C%26%20%7D%20n%3D4%5C%5C%5C%5C%26%5CDelta%20a%3D%5Cfrac%7Bb-a%7D%7Bn%7D%3D%5Cfrac%7B%5Cpi%20%2F%202-%5Cpi%20%2F%208%7D%7B9%7D%3D%5Cfrac%7B3%20%5Cpi%7D%7B32%7D%5C%5C%5C%5C%26x_%7B0%7D%3D%5Cpi%20%2F%208%2C%20x_%7B1%7D%3D%5Cfrac%7B%5Cpi%7D%7B8%7D%2B%5Cfrac%7B3%20%5Cpi%7D%7B32%7D%3D%5Cfrac%7B7%20%5Cpi%7D%7B32%7D%5C%5C%5C%5C%26x_%7B2%7D%3D%5Cfrac%7B5%20%5Cpi%7D%7B16%7D%2C%20%5Cquad%20x_%7B3%7D%3D%5Cfrac%7B13%20%5Cpi%7D%7B32%7D%2C%20x_%7B4%7D%3D%5Cpi%20%2F%202%5C%5C%5C%5C%26f%5Cleft%28x_%7B0%7D%5Cright%29%3Df%281%20%2F%208%29%3D0.8535%5C%5C)
![&f\left(x_{1}\right)=f\left(\frac{7 \pi}{32}\right)=0.5975\\\\\&f\left(x_{2}\right)=f\left(\frac{5 \pi}{16}\right)=0.3086\\\\\&f\left(a_{3}\right)=f\left(\frac{13 \pi}{32}\right)=0.0842\\\\\&L_{4}=\sum_{k_{0}^{=0}}^{3} f\left(x_{k}\right) \Delta x\\\\&=\Delta \\\\x\left[f\left(x_{0}\right)+f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)\right]\\\\\&=\frac{3 \pi}{32}[0.8535+0.5975+0.3056+0.0842]\\\\&L_{4}=0.5431](https://tex.z-dn.net/?f=%26f%5Cleft%28x_%7B1%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B7%20%5Cpi%7D%7B32%7D%5Cright%29%3D0.5975%5C%5C%5C%5C%5C%26f%5Cleft%28x_%7B2%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B5%20%5Cpi%7D%7B16%7D%5Cright%29%3D0.3086%5C%5C%5C%5C%5C%26f%5Cleft%28a_%7B3%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B13%20%5Cpi%7D%7B32%7D%5Cright%29%3D0.0842%5C%5C%5C%5C%5C%26L_%7B4%7D%3D%5Csum_%7Bk_%7B0%7D%5E%7B%3D0%7D%7D%5E%7B3%7D%20f%5Cleft%28x_%7Bk%7D%5Cright%29%20%5CDelta%20x%5C%5C%5C%5C%26%3D%5CDelta%20%5C%5C%5C%5Cx%5Cleft%5Bf%5Cleft%28x_%7B0%7D%5Cright%29%2Bf%5Cleft%28x_%7B1%7D%5Cright%29%2Bf%5Cleft%28x_%7B2%7D%5Cright%29%2Bf%5Cleft%28x_%7B3%7D%5Cright%29%5Cright%5D%5C%5C%5C%5C%5C%26%3D%5Cfrac%7B3%20%5Cpi%7D%7B32%7D%5B0.8535%2B0.5975%2B0.3056%2B0.0842%5D%5C%5C%5C%5C%26L_%7B4%7D%3D0.5431)
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The complete Question is attached below
f(x) = log2 x
f(40) = log2 40
40 = 2^y
2^5 = 32 and 2^6 = 64
so f(40) lies between integers 5 and 6.
Answer:
64.3 m²
Step-by-step explanation:
The figure is composed of a square and a quarter circle.
Therefore, area of the figure = area of the square + area of quarter circle
= s² + ¼(πr²)
Where,
s = 6 m
r = 6 m
Plug in the values
Area = 6² + ¼(π*6²)
Area = 64.274334 ≈ 64.3 m² (nearest tenth)
Answer:
1/14
Step-by-step explanation:
dont really know how to explain... just make sure to change both to a common denominator
Answer:
5/28
Step-by-step explanation:
First of all, probability is defined as the number of ways a certain event can happen divided by the total ways of an event happening. In this scenario, we are asked to find the number of non-defective computers divided by the total ways that a sample of two can be chosen.
The number of ways to choose 2 computers from 8 can be written as
, which is equal to
, which is 28. Now, there are 3 defective computers, for a total of 5 non-defective computers, so the probability is 5/28.