Question

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.
-3
0
-16/5
8/3
-2
7/2

Answer 1

Answer:

• 8/3
• -2

Step-by-step explanation:

The integral from a to 4 is ...

$\int\limits^4_a {(3x-1)} \, dx =\dfrac{3}{2} (4^2-a^2)-(4-a) = 20+a-\dfrac{3}{2}a^2$

We want the value of this to be 12, so we can write ...

  -3/2a^2 +a +20 = 12

  3a^2 -2a -16 = 0 . . . . . subtract the left side and multiply by 2

  (3a -8)(a +2) = 0 . . . . . factor

  a = 8/3 or -2  . . . . . . . .values of a that make the factors zero

The only possible values of a are 8/3 and -2.