 As a general rule, arbitrary functions don't have elementary antiderivatives, so what happens when we have to find a definite integral for such a function? Remember that a problem exists whether or not you can solve it, so we have to find something to do, and in this case we must rely on some numerical integration technique. So remember that the definite integral can be defined as the limit of the Riemann sum as the width of the widest interval goes to zero. Now previously we found Riemann sums by using right endpoints, left endpoints, lower rectangles, and upper rectangles. And all of these were rectangles and we just had to decide where we wanted to take the height, and so we'll take a radical step and choose the midpoint. So let's find the approximate value of this integral and we'll use the midpoint and n equals two intervals. So let's go ahead and graph this. If we have n equals two equal width intervals, the intervals will run from x equals zero to x equals two, and from x equals two to x equals four. So in the first interval, the midpoint will be at x equals one, and so we'll find the coordinates of the point on the graph at x equals one. In the second interval, the midpoint will be at x equals three, and again we'll find the coordinates of the point on the graph at x equals three. Now the midpoints give us the tops of two rectangles, and so we'll find the areas of the two rectangles, and if we add these areas up that will give us an approximation to the definite integral. Now we can derive a formula for the area given by the midpoint approximation. We won't actually do that. We'll leave that as an exercise for the viewer. But now let's do something completely different. We can take a radical step and not use rectangles. Instead, we can use trapezoids. So for this, it's helpful to remember that the area of a trapezoid with bases b1 and b2 at height h will be given by the formula, and you can think about this as the height times the average of the two bases. So again, let's use n equals two rectangles to approximate the same integral, and so our trapezoid will actually join the corners of our region. So the first trapezoid will have one corner at 0, 1, and the other at 2, root 17. So its area will be the second trapezoid will have corner at 2, root 17, and the other at 4, root 257. And so its area will be, and so the trapezoidal approximation will be the sum of these areas. Now we could derive a formula for the trapezoidal approximation. We shouldn't, but we will. So suppose we partition our interval at points x0, x1, x2, and so on, where the partitions have equal width delta x. So we need to know the heights of each of these points, so we find their coordinates. Now the first trapezoid will have bases fx0 and fx1, so the area will be, but since we've assumed the intervals have equal width, x1 minus x0 is just delta x, and so the area of the first trapezoid will be. Similarly the second trapezoid will have area, and again since we've assumed that the intervals have equal width, the area of the second trapezoid will be. And similarly we can find the area of the third trapezoid, and so on, all the way up to the last trapezoid. Now since every one of these areas has a one-half delta x in common, we can factor it out and get the sum of the components. And the thing that's worth recognizing here is that except for the first and last function values, every other function value appears twice. And so this sum can be written as, and this gives us a nice formula for the trapezoid approximation. And so we can approximate this definite integral using the trapezoidal approximation with n equals 4 equal width partitions. So with n equals 4 partitions, the interval between 0 and 4 will have the partition points, x equals 0, 1, 2, 3, and 4. We find our function values, and we can read the trapezoidal approximation as one-half the sum of all the function values, doubling those that are inside the interval. In other words, not the first, not the last, but all the others. And so we find our trapezoidal approximation.