202210191139 Solution to 1973-CE-AMATH-II-XX

Find the area bounded by the curve $y=4x-x^2$ and the straight line $x-y+2=0$ .

Roughwork.

Rewriting,

$\begin{cases} C:y=-x^2+4x \\ L:y=x+2 \\ \end{cases}$

Solving,

$\begin{aligned} x+2 & = -x^2+4x \\ 0 & = x^2-3x+2 \\ 0 & = (x-1)(x-2) \\ x & = 1,2 \\ \end{aligned}$

$\begin{aligned} x=1 \Rightarrow y=(1)+2=3 \\ x=2 \Rightarrow y=(2)+2=4 \\ \end{aligned}$

Now that the points of intersection are $(1,3)$ and $(2,4)$ , plot a graph below.

By either definite integral,

$\textrm{Area} & = \begin{cases} \displaystyle{\int_{1}^{2}\big(y_C(x)-y_L(x)\big)\,\mathrm{d}x }\\ \displaystyle{\int_{3}^{4}\big(x_L(y)-x_C(y)\big)\,\mathrm{d}y }\\ \end{cases}$

The upper integral is easier solvable than the lower one. Exercise.

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31