Find the area of a trapezoid using integration

In summary: There is no equation with that notation in the picture.In summary, the homework equation is y= (1/2)x+ 3. Setting x= -2, y= 2, gives the equation 2= a(-2)+ b. Setting x= 4, y= 5, gives 5= a(4)+ b. Subtracting the first equation from the second, (4a-(-2a))+ (b- b)= 5- 2 or 6a= 3 so a= 3/6= 1/2. Putting this into 4a+ b= 5 gives 2+ b= 5 so b= 3.
  • #1
bobsmith76
336
0

Homework Statement


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Homework Equations



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The Attempt at a Solution



I don't see why you have to put the ((x/2)+ 3)dx next to the equation for a trapezoid. I also don't understand what number would equal x, 4? 2?. Also, the dx, what equation am I supposed to be deriving? .5((B+b)/2)h? that equation doesn't derive.
 
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  • #2
The left edge of the trapezoid goes from (-2, 2), on the left, to (4, 5), on the right. Since it is a line, its equation is of the form y= ax+ b. Setting x= -2, y= 2, that gives the equation 2= a(-2)+ b or -2a+ b= 2. Setting x= 4, y= 5, that gives 5= a(4)+ b or 4a+ b=5, two equations to solve for b. Subtracting the first equation from the second, (4a-(-2a))+ (b- b)= 5- 2 or 6a= 3 so a= 3/6= 1/2. Putting this into 4a+ b= 5 gives 2+ b= 5 so b= 3. That is, the equation of the top line is y= (1/2)x+ 3 which is the same as the f(x)=(x/2)+ 3 shown in the picture.

Now, imagine drawing a thin rectangle from the base, y= 0 to the top line y= f(x)= (x/2)+ 3. It height is just the y value minus 0, (x/2)+ 3. We can think of the base as being the very small number [itex]\delta x[/itex]. The area of that rectangle, height times base, is [itex]((x/2)+ 3)\Delta x[/itex] and, since we can cover the trapezoid by such rectangles, we can approximate the area by [itex]\sum ((x/2)+ 3)\Delta x[/itex]. That's what is called a "Riemann sum". We can take the limit as the bases get smaller and smaller, converting that Riemann sum to the integral [itex]\int ((1/2)x+ 3)dx[/itex]. Of course, x ranges from -2 on the left to 4 on the right so the integral is [itex]\int_{-2}^4 ((1/2)x+ 3)dx[/itex].

If F'(x)= f, that is, if f is the derivative of F (F is the anti-derivative of f) then [itex]\int_a^b f(x)dx= F(b)- F(a). You evaluate the anti-derivative at the lower and upper values, -2 and 4 in this problem and subtract.
 
  • #3
I found this technique over at the Khan Academy

1. find the antiderivative
2. subtract the value of the B antiderivative from the A antiderivative

Hopefully I'll understand why later.

By the way, how do you get those equations with sigma notation in your post.
 

FAQ: Find the area of a trapezoid using integration

What is the formula to find the area of a trapezoid using integration?

To find the area (\(A\)) of a trapezoid using integration, you can use the following formula:

\[A = \int_{a}^{b} f(x) \, dx\]

Where:

  • \(a\) and \(b\) are the x-coordinates of the two parallel sides of the trapezoid.
  • \(f(x)\) represents the function that defines the height of the trapezoid at each point \(x\) between \(a\) and \(b\).

This formula calculates the area by integrating the height function \(f(x)\) with respect to \(x\) over the interval \([a, b]\).

How do you use this formula to find the area of a trapezoid?

To find the area of a trapezoid using integration, follow these steps:

  1. Identify the trapezoid: Determine the trapezoid's characteristics, including the lengths of its two parallel sides (the longer side and the shorter side) and the function that defines its height.
  2. Choose a coordinate system: Select an appropriate coordinate system for your problem. Decide which side of the trapezoid corresponds to the x-axis.
  3. Express the height function \(f(x)\): Write an expression for the height of the trapezoid as a function of \(x\). This function represents how the height changes along the x-axis.
  4. Determine the integration limits: Identify the values of \(a\) and \(b\), which are the x-coordinates of the two endpoints of the interval over which you want to find the area. These values correspond to the left and right ends of the trapezoid's base.
  5. Set up the integration: Use the formula \(A = \int_{a}^{b} f(x) \, dx\) to set up the integration. Replace \(a\), \(b\), and \(f(x)\) with the values determined in the previous steps.
  6. Calculate the definite integral: Evaluate the definite integral to find the area \(A\). This involves finding the antiderivative of \(f(x)\) and applying the Fundamental Theorem of Calculus.
  7. Obtain the area: The result of the definite integral is the area of the trapezoid enclosed by the specified x-coordinates and the height function \(f(x)\).

Keep in mind that the choice of coordinate system, the direction of the x-axis, and the function \(f(x)\) may vary depending on how the trapezoid is oriented in space. Ensure that your setup aligns with the problem at hand.

Can you provide an example of finding the area of a trapezoid using integration?

Sure! Let's find the area of a trapezoid with bases of length \(a = 2\) and \(b = 6\) units and a height function \(f(x) = 3x + 1\).

  1. Identify the trapezoid: We have a trapezoid with bases of length \(a = 2\) and \(b = 6\) units, and the height function is \(f(x) = 3x + 1\).
  2. Choose a coordinate system: We'll use a standard Cartesian coordinate system with the x-axis aligned along the longer base of the trapezoid.
  3. Express the height function \(f(x)\): The height function is \(f(x) = 3x + 1\).
  4. Determine the integration limits: Our integration limits are \(a = 2\) and \(b = 6\), which correspond to the x-coordinates of the two bases.
  5. Set up the integration: We set up the integration as follows: \[A = \int_{2}^{6} (3x + 1) \, dx\]
  6. Calculate the definite integral: Evaluate the definite integral: \[A = \left[\frac{3}{2}x^2 + x\right]_{2}^{6} = \left(\frac{3}{2}(6)^2 + 6\right) - \left(\frac{3}{2}(2)^2 + 2\right) = 45\] square units.
  7. Obtain the area: The area of the trapezoid is \(A = 45\) square units.

So, the area of the trapezoid is 45 square units.

Are there any variations of the formula for different types of trapezoids?

The formula \(A = \int_{a}^{b} f(x) \, dx\) can be adapted for various types of trapezoids, including those with inclined bases or non-standard orientations. The key is to properly define the height function \(f(x)\) and determine the appropriate integration limits \(a\) and \(b\) based on the specific characteristics of the trapezoid.

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