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Integrate each function by using the table in Appendix $D$.A good representation of the cables between towers of the $2280-\mathrm{m}$ section of the Golden Gate Bridge is $y=0.000370 x^{2}$ for $-1140 \leq x \leq 1140,$ where $x$ and $y$ are in meters. Find the length of the cables (see Exercise 35 of Section 26.6 ).

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 11

Integration by Use of Tables

Integrals

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Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

01:57

Golden Gate cables The pro…

01:06

The profile of the cables …

03:01

Length of a Cable An elect…

01:36

Solve each problem. Th…

02:13

A cable of the Golden Gate…

03:10

Suspension Bridge A cable …

01:03

The Golden Gate Bridge is …

If an $x$ -axis is superim…

06:41

APPLICATIONSIf an $x$ …

02:05

Solve each problem.The…

So here on this problem, we need to find the length of this cable. And we are also given that this cable can be represented by an equation which is why is equals two 0.0 370 X. Square. And they were also given that X. Is greater than or equal to negative 1140 less than or equals to 1140 Now this is the input data. Now basically we have to calculate the length of the ark. So the length of the ark is of a cough from X. Is equal to a two X equals two. B. Is given by the formula as equals to integration A to B under route one plus. Do you buy over the X whole square dot dx. So this this basically can be treated as a cove and we can find the length of the scope by using this formula. Let me first find this dy over the X. So this is why we are giving us this much. So we can differentiate it. We can differentiate this. So do you buy over the X. Will be equals to two times 0.0 370 X. And it will be equal to 0.0 074 X. Now you can plug it here and if we have this negative 11 this one is A basically and this is B. So we can use this formula to evaluate this. So it will be cool too. This is equal integration. If we have a negative or 1140 to be is 1140 under roof. One plus dy over dx square. So this is this much zero point 00074 X. And it's square. Since this is an even function. So we can use the integral of properties and it can also be written us two times 0 to 1140 Under route one plus 0.74 X. And its square dot dx. Now we're going to evaluate this integral to find the length. So we're going to use one formula which is the integration under route one plus X. Cold square. Not dx is equal to X over two. Under rule one plus a square square less one over to a natural log. This is a X plus. Underwrote one plus a square X. Square. So I'm going to use this formula to evaluate this integral. If you compare this formula with this integral, we can say that a we have 0.74 X. So you can plug it all this value here. So X by two as it is on the route or one plus ex a square expert. So it would be 0.74 X whole square. We can write this. We're also plus one over to a natural look, if we have a 0.0 746 plus underwrote one plus a square X squared. So this is 0.74 X. And it's squared. Now we have all the limits. So limit is 02 1140 And this too as it is. So it will be we can write it here. This is too as it is. Now, can you evaluate this into this expression by putting up in the limits upper limit minus the lower limit. And after putting the limits, we'll get the final answer to be 2527 0.7 Which can be approximated to 2513 m. So this is the length of the cable. So I hope you have understood the problem. Thank you.

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