## Calculus: Early Transcendentals (2nd Edition)

(a) $$A(2)=\frac{8+4}{2}\cdot2=12.$$ (b) $$A(6)=A_1+A_2=9+12=21.$$ (c) $$A(x)=\Bigg\{_{2x+9,\quad x>3}^{-x^2+8x,\quad x\leq3}.$$
(a) In this case, we will have a right trapezoid with the vertical sides of $8$, where the graph of $f$ meets the $y$ axis, and $f(2)=-2\cdot2+8=4$ where the vertical line $t=2$ meets the graph of the function. The area of such a trapezoid is equal to the arithmetic mean of the length of its sides times the "height" i.e. the distance between them which is equal to $2$: $$A(2)=\frac{8+4}{2}\cdot2=12.$$ (b) We can split this region to the triangle bounded by the graph of the function for $t$ between $0$ and $3$, the $y$ axis and the horizontal line $y=f(3)=-2\cdot3+8=2$. This is the right triangle with the legs of $8-2=6$ and $3$ so this portion of the area is $$A_1=\frac{1}{2}\cdot6\cdot3=9.$$ The other part is the rectangle bounded by the horizontal line $y=2$, the $t$ axis, the $y$ axis and the vertical line $t=6$. This rectangle has the sides of $6$ and $2$ so this portion of the area is $$A_2=6\cdot 2=12.$$ The total area is $$A(6)=A_1+A_2=9+12=21.$$ (c) We have to split this into cases. If $x\leq 3$ then the region will be the trapezoid bounded by the $y$ axis, the $t$ axis, the graph of $f$ and the vertical line $t=x$ so its bases will be $8$ and $f(x)=-2x+8$ while the height is $x$ which would give the area of $$A(x)=\frac{8-2x+8}{2}\cdot x=-x^2+8x.$$ If $x>3$ then we will calculate the joint area of $A(3)$ i.e. the trapezoid bounded by $y$ and $t$ axes, the graph of $f$ and the vertical line $t=3$ and the area of the rectangle bounded by $y=2$, the $t$ axis and vertical lines $t=3$ and $t=x$: $$A(x)=A(3)+2\cdot (x-3)=-3^2+8\cdot3+2x-6=2x+9.$$ The final answer is $$A(x)=\Bigg\{_{2x+9,\quad x>3}^{-x^2+8x,\quad x\leq3}.$$