Definite Integrals Exercises

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.

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.

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.

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.

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 and calculating 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.