Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the area of a trapezoid using integration

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. 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.
  2. jcsd
  3. Jan 31, 2012 #2


    User Avatar
    Science Advisor

    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.
  4. Jan 31, 2012 #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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook