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The table shows values of a force function $ f(x) $, where x is measured in meters and $ f(x) $ in newtons. Use Simpson's Rule to estimate the work done by the force in moving an object a distance of 18 m.
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 7
Approximate Integration
Integration Techniques
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University of Michigan - Ann Arbor
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Lectures
01:53
In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.
27:53
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
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The table shows values of …
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Work from force How much w…
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Shown is the graph of a fo…
Okay, so this question wants us to estimate the work done, which is given by the integral under a force distance graph. So it wants us to use Simpson's role. So to figure out how many sub intervals we need let's look at how many data points were given. So were given seven data points. So our end is data points minus one. So it's six and then to find our Delta X be minus a over end, which is, well, we're ending at 18 and we're starting at zero, and we got six some intervals, so Delta X should be three. So now we get S of six is approximately equal to the world. So s of six is Delta X over three, which is three divided by three. So it's just one times f of zero plus four times f of three plus two times f of six plus all the way up to s of 18. And then just using the values that we know from the table, we can see that the work is approximately 148 Jules and again, all we do is plugging the values from the table injury to the efs remembering the alternate, the fours and the twos. Except for the end points, of course, and we'll get an answer of 1 48 Jules.
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