Problem Two:

In this problem, we are asked to find the area of an irregular trapezoid. The trapezoid has a vertex at the origin, "x-equals-zero…y-equals-zero." From the origin, one edge extends to the point, "x-equals-seven…y-equals-fifteen." From this point, another edge extends horizontally to the right for six units. This edge terminates at the point, "x-equals-thirteen…y-equals-fifteen. From the point, "thirteen…fifteen," a third edge extends vertically downward for fifteen units, terminating at the point, "x-equals-thirteen…y-equals-zero." The final edge extends from this point back to the origin along the x-axis.
The figure may be divided into two parts. The first part is a right triangle with one acute vertex at the origin, a second acute vertex at the point, "seven…fifteen," and the right angle at the point, "zero…seven." The second part of the figure is a rectangle with a height of seven units, and a base extending along the x-axis from the point, "zero…seven," to the point, "zero…thirteen."
The right triangle has a base equal to seven units and a height equal to fifteen units. The area of a triangle is one-half of the product of its base and its height. One half of seven units times fifteen units is fifty-two-and-one-half square units. The rectangle has a base equal to six units and a height equal to fifteen units. The area of a rectangle is the product of its base and its height. Six units times fifteen units equals ninety square units. The total area is the sum of the parts. "Fifty-two-and-one-half square units," plus, "ninety square units," equals, "one-hundred-and forty-two-and-one-half-square-units."
Expressing all units in feet, the area is one-hundred-and forty-two-and-one-half-square-feet.
Expressed as an integral, the area is again the sum of two parts. The first part is the area between the x-axis and the line from the origin to the point, "seven…fifteen." The equation of this line is, "y equals fifteen-over-seven x." The corresponding integral is, "The integral from x-equals-zero to x-equals-seven of fifteen-over-seven-x-dee-x." The value of this integral, again using feet for our unit of measure, is fifty-two-and-one-half square feet. The second part is the area between the horizontal line y-equals-fifteen and the x-axis between x equals seven and x equals thirteen. The corresponding integral is, "The integral from x-equals-seven to x-equals-thirteen of fifteen-dee-x." The value of this integral is ninety square feet. Again, the area is the sum of the two integrals, or one-hundred-and-forty-two-and-one-half square feet.