Constant and a variable in a squareroot needs integrating

In summary, the equation is required to be integrated to apply the trapezium rule, but the x(20-x)^1/2 is confusing me. I attempted the problem and got a resultant that doesn't seem to be correct. From what I can tell, you'd be greatly helped by the substitution u=20-x, du=-dx. Then rewrite x as 20-u to get -(20 - u) u1/2 du. If you read the problem statement carefully, you are instructed first to fill out the missing depth values in the table. Did you do that?
  • #1
ferrelhadley
9
0

Homework Statement



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http://s39.photobucket.com/albums/e178/dorlomin/enviroment/?action=view&current=core2.jpg"

Homework Equations


In the question shown the equation is required to be integrated to apply the trapezium rule. However the x(20-x)^1/2 is confusing me somewhat.


The Attempt at a Solution



When I attempt the question the only way I can think of doing it is be multiplying the 'x' onto the '(20-x)^1/2'
But the resultant I get is
(4.47x - x^3/2)

The get 4.47/2x^2 - 2/5x^5/2

Can anyone point me to a worked example of how to deal with a constant and a variable in a squareroot?

My solutions is not giving me the right answers...
 
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  • #2
From what I can tell, you'd be greatly helped by the substitution u=20-x, du=-dx. Then rewrite x as 20-u to get -(20 - u) u1/2 du.
 
  • #3
If you read the problem statement carefully, you are instructed first to fill out the missing depth values in the table. Did you do that?

Once the table is complete, then you are to calculate the cross sectional area of the river channel by using the trapezoidal rule for numerical integration. Do you understand what the trapezoidal rule is and how it is applied?

Once you have calculated the cross sectional area, you are given a current velocity for the water flowing in the channel and asked to estimate the flow rate of the water.

The problem does not ask you to integrate the depth function analytically to find the cross sectional area of the channel. Instead, you are supposed to calculate area numerically using the trapezoidal rule.
 
  • #4
SteamKing said:
Do you understand what the trapezoidal rule is and how it is applied?

I did not but have now found out and found the missing values
1.6 3.2 and applied the trapezoid formula and gotten 43.86

Many thanks
 

1. What is a constant in a square root?

A constant in a square root is a fixed numerical value that does not change when the square root is evaluated. It can also be referred to as a coefficient or a factor.

2. How is a variable used in a square root?

A variable in a square root is a symbol that represents a quantity that can change or vary. It is typically represented by letters such as x, y, or z and is used to find the value of the square root.

3. What does it mean to integrate a square root?

To integrate a square root means to find the antiderivative or the reverse operation of differentiation. It is a mathematical process used to find the original function when the derivative is known.

4. Why is integrating a square root important?

Integrating a square root is important because it allows us to solve complex mathematical problems and model real-life situations. It is also a fundamental concept in calculus and is used in various fields such as physics, engineering, and economics.

5. How do you integrate a square root?

To integrate a square root, you can use various techniques such as substitution, integration by parts, or the method of partial fractions. It is important to understand the properties and rules of integration and choose the appropriate method for the given problem.

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