## Calculus: Early Transcendentals (2nd Edition)

(a) $$A(2)=\frac{1}{2}\cdot1\cdot2=1.$$ (b) $$A(6)=\frac{1}{2}\cdot3\cdot6=9.$$ (c) $$A(x)=\frac{1}{2}\cdot \frac{x}{2}\cdot x=\frac{x^2}{4}.$$
(a) In this case, we will have the right triangle bounded by $f(t)$, the $t$ axis and the vertical line $t=2$. This vertical line will intercept $f(t)$ at the point $(2,f(2))=(2,2/2)=(2,1)$. This means that one leg of this triangle is equal to $2$ (horizontal one), and the other is equal to $1$ (the vertical one), so we have for the area: $$A(2)=\frac{1}{2}\cdot1\cdot2=1.$$ (b) In this case, we will have the right triangle bounded by $f(t)$, the $t$ axis and the vertical line $t=6$. This vertical line will intercept $f(t)$ at the point $(6,f(6))=(6,6/2)=(6,3)$. This means that one leg of this triangle is equal to $6$ (horizontal one), and the other is equal to $3$ (the vertical one), so we have for the area: $$A(6)=\frac{1}{2}\cdot3\cdot6=9.$$ (c) In this case, we will have the right triangle bounded by $f(t)$, the $t$ axis and the vertical line $t=x$. This vertical line will intercept $f(t)$ at the point $(x,f(x))=(x,x/2)$. This means that one leg of this triangle is equal to $x$ (horizontal one), and the other is equal to $x/2$ (the vertical one), so we have for the area: $$A(x)=\frac{1}{2}\cdot \frac{x}{2}\cdot x=\frac{x^2}{4}.$$