if the area (in square units) of the region under the curve of the function f(x) = 3x − 1 on the interval [a, 4], where a < 4, is 12 square units, identify all the possible values of a.

  -2, 8/3

Step-by-step explanation:

You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...

  A = (1/2)(f(a) +f(4))(4 -a)

  = (1/2)((3a -1) +(3·4 -1))(4 -a)

  = (1/2)(3a +10)(4 -a)

We want this area to be 12, so we can substitute that value for A and solve for "a".

  12 = (1/2)(3a +10)(4 -a)

 24 = (3a +10)(4 -a) = -3a² +2a +40

  3a² -2a -16 = 0 . . . . . . subtract the right side

  (3a -8)(a +2) = 0 . . . . . factor

Values of "a" that make these factors zero are ...

  a = 8/3, a = -2

The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.

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Alternate solution

The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.

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step-by-step explanation: