Answer:
16.5 square units
Step-by-step explanation:
You are expected to integrate the function between x=1 and x=4:
[tex]\displaystyle\text{area}=\int_1^4{(5x-x^2)}\,dx=\left.\left(\dfrac{5}{2}x^2-\dfrac{1}{3}x^3\right)\right|_{x=1}^{x=4}\\\\=\dfrac{5(4^2-1^2)}{2}-\dfrac{4^3-1^3}{3}=37.5-21=\boxed{16.5}[/tex]
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Additional comment
If you're aware that the area inside a (symmetrical) parabola is 2/3 of the area of the enclosing rectangle, you can compute the desired area as follows.
The parabolic curve is 4-1 = 3 units wide between x=1 and x=4. It extends upward 2.25 units from y=4 to y=6.25, so the enclosing rectangle is 3×2.25 = 6.75 square units. 2/3 of that area is (2/3)(6.75) = 4.5 square units.
This region sits on top of a rectangle 3 units wide and 4 units high, so the total area under the parabolic curve is ...
area = 4.5 +3×4 = 16.5 . . . square units
