By Euler's method the <em>numerical approximate</em> solution of the <em>definite</em> integral is 4.189 648.
<h3>How to estimate a definite integral by numerical methods</h3>
In this problem we must make use of Euler's method to estimate the upper bound of a <em>definite</em> integral. Euler's method is a <em>multi-step</em> method, related to Runge-Kutta methods, used to estimate <em>integral</em> values numerically. By integral theorems of calculus we know that definite integrals are defined as follows:
∫ f(x) dx = F(b) - F(a) (1)
The steps of Euler's method are summarized below:
- Define the function seen in the statement by the label f(x₀, y₀).
- Determine the different variables by the following formulas:
xₙ₊₁ = xₙ + (n + 1) · Δx (2)
yₙ₊₁ = yₙ + Δx · f(xₙ, yₙ) (3) - Find the integral.
The table for x, f(xₙ, yₙ) and y is shown in the image attached below. By direct subtraction we find that the <em>numerical</em> approximation of the <em>definite</em> integral is:
y(4) ≈ 4.189 648 - 0
y(4) ≈ 4.189 648
By Euler's method the <em>numerical approximate</em> solution of the <em>definite</em> integral is 4.189 648.
To learn more on Euler's method: brainly.com/question/16807646
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<em>Answer:</em>
<em>The net speed per 24 hours is 5 - 4 = 1 foot. </em>
<em>1 ft / 24 hrs = 16 ft / x hrs </em>
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<em>16 * 24 hrs = 16 days (384 hrs)
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Answer:
1
Step-by-step explanation:
To get the GCF of 44 and 47, factor each value,then we choose copies of factors and multiply them.
Solution:
A function is always a relation but a relation is not always a fucntion.
For example
we can make a realtion of student roll number and their marks obtained in mathematics.
So we can have pairs like (a,b), (c,d)..etc.
Its a realtion but it may not be function. Because function follows that for same input there should not be diffrent output, aslo there could be many inputs to one output in the case of constant function . But this doesn't holds a necessary condition in case of relation.
Because two diffrent students with two diffrent Roll number may have same marks.
Hence the foolowing options holds True in case of a function.
A) many inputs to many outputs or one input to one output.
D) one input to one output or many inputs to one output.