Answer:
0.5 square units
Step-by-step explanation:
Find the equation of the parabola in the form [tex]y=ax^2.[/tex] This parabola passes through the point (3,2), so
[tex]2=a\cdot 3^2\\ \\2=9a\\ \\a=\dfrac{2}{9}[/tex]
Thus, the parabola equation is
[tex]y=\dfrac{2}{9}x^2[/tex]
Now, find the equation of the tangent line.
[tex]y'=\left(\dfrac{2}{9}x^2\right)'=\dfrac{2}{9}\cdot 2x=\dfrac{4}{9}x\\ \\y'(3)=\dfrac{4}{9}\cdot 3=\dfrac{4}{3}[/tex]
The equation of the tangent line is
[tex]y-y(x_0)=y'(x_0)(x-x_0)\\ \\y-2=\dfrac{4}{3}(x-3)\\ \\y=\dfrac{4}{3}x-2[/tex]
This line intersects x-axis at point (1.5, 0), because
[tex]\dfrac{4}{3}x-2=0\\ \\x=2\cdot \dfrac{3}{4}=1.5[/tex]
Find the shaded area:
[tex]A=\int\limits_0^{1.5}\dfrac{2}{9}x^2dx+\int\limits_{1.5}^3\left(\dfrac{2}{9}x^2-\dfrac{4}{3}x+2\right)dx=\left(\dfrac{2}{9}\cdot \dfrac{x^3}{3}\right)|_0^{1.5}+\left(\dfrac{2}{9}\cdot \dfrac{x^3}{3}-\dfrac{4}{3}\cdot \dfrac{x^2}{2}+2x\right)|_{1.5}^3=\\ \\=\dfrac{2}{27}\cdot (1.5^3-0^3)+\dfrac{2}{27}\cdot (3^3-(1.5)^3)-\dfrac{2}{3}\cdot(3^2-1.5^2)+2\cdot (3-1.5)=[/tex]
[tex]=\dfrac{2}{27}\cdot \dfrac{27}{8}+\dfrac{2}{27}\cdot \left(27-\dfrac{27}{8}\right)-\dfrac{2}{3}\cdot \left(9-\dfrac{9}{4}\right)+2\cdot \dfrac{3}{2}=\\ \\=\dfrac{1}{4}+2-\dfrac{1}{4}-6+\dfrac{3}{2}+3=\\ \\=0.5\ un^2.[/tex]
