For the integral, (a) find the indicated approximation, and (b) determine if your approximation is an under-estimate, an over-estimate, or exact.
RHS(3) for
Answer
We divide the interval [-5, 1] into 3 sub-intervals of length 2. On each sub-interval we use the value of f at the right endpoint as the "height" of the rectangle.
Since the function we're integrating is increasing, a right-hand sum is an over-estimate.
Example 2
For the integral, (a) find the indicated approximation, and (b) determine if your approximation is an under-estimate, an over-estimate, or exact.
LHS(5) for
Answer
We divide the interval [-10, 0] into 5 sub-intervals of length 2. On each sub-interval we use the value of f at the left endpoint as the "height".
Since the function we're integrating is decreasing, a left-hand sum is an overestimate.
Example 3
For the integral, (a) find the indicated approximation, and (b) determine if your approximation is an under-estimate, an over-estimate, or exact.
MID(3) for
Answer
We split the interval [-4,2] into sub-intervals [-4, -2], [-2, 0], and [0, 2], which have midpoints x = -3, -1, and 1, respectively. All sub-intervals have length 2.
Since the function we're integrating is concave up, the midpoint sum is an underestimate.
Example 4
For the integral, (a) find the indicated approximation, and (b) determine if your approximation is an under-estimate, an over-estimate, or exact.
TRAP(5) for
Answer
Divide the interval [-10, 0] into 5 sub-intervals of length 2. We find the value of f at each endpoint and draw the trapezoids. To find the actual trapezoid sum, we can use our handy formula.
Since the function we're integrating is concave down, the trapezoid sum is an underestimate.
Example 5
For the integral, (a) find the indicated approximation, and (b) determine if your approximation is an under-estimate, an over-estimate, or exact.
TRAP(4) for
Answer
This time we get sub-intervals of length 5.
Since a straight line has no concavity, the trapezoid rule gives the exact value of the integral. We can verify this by finding the areas of the triangles above and below the x-axis andcalculating the value of the integral exactly.
The line crosses the x-axis when x = 6. The big triangle has base 16 and "height" f (-10) = -8, so its weighted area is -64.
The little triangle has base 4 and height f (10) = 2, so its weighted area is 4. Combining these, we get
which confirms that the trapezoid rule gave us the exact value of the integral.
crosses the x-axis when x = 6. The big triangle has base 16 and "height" f (-10) = -8, so its weighted area is -64.