Area of the region enclosed by $y=x^2-1$, $y=2x-1$, and $y=x+1$

Question

Find the area of the region enclosed by $y=x^2-1$, $y=2x-1$, and $y=x+1$.

Like the question I asked before, I still do not understand how to find these areas.

Answer 1

Find the intersection of $y=x^2-1$ and $y=x+1$.

$x^2-x-2=0$

$x=-1,2$

$$S1=\int_{-1}^2\left[x+1-\left(x^2-1\right)\right]dx$$

Find the intersection of $y=x^2-1$ and $y=2x-1$

$x^2-2x=0$

$x=0,2$

$$S2=\int_0^2\left[2x-1-\left(x^2-1\right)\right]dx$$

What does $S1-S2$ give you?

fast answer posted due to desmos 'misc' function.

Answer 2

This can be done using addition also.

Find the intersection of $y=2x-1$ and $y=x^2-1$

$x=0,2$

$$S1=\int_{-1}^0\left[x+1-\left(x^2-1\right)\right]dx$$

Find the intersection of $y=x^2-1$ and $y=x+1$

$x=-1,2$

$$S2=\int_0^2\left[x+1-\left(2x-1\right)\right]dx$$

So, what can be done with $S1$ and $S2$?

Answer 3

\begin{align} f_1(x)&=x^2-1 ,\quad f_2(x)=2x-1 ,\quad f_3(x)=x+1 ,\\ A&=(-1,0),\quad B=(0,-1),\quad C=(2,3),\quad D=(0,1) . \end{align}

\begin{align} [ABCD]&=[ABD]+[BCD] \\ &= \int_{-1}^{0} \int_{f_1(x)}^{f_3(x)} 1\cdot\,dy \,dx +\int_{0}^{2} \int_{f_2(x)}^{f_3(x)} 1\cdot\,dy \,dx \\ &= \int_{-1}^{0} \int_{x^2-1}^{x+1} 1\cdot\,dy \,dx +\int_{0}^{2} \int_{2x-1}^{x+1} 1\cdot\,dy \,dx \\ &=\int_{-1}^0 (x+2-x^2) \,dx +\int_0^2 (2-x)\,dx \\ &=\left(\left.\tfrac12\,x^2+2x-\tfrac13\,x^3\right)\right|_{-1}^0 +\left(\left. 2x-\tfrac12\,x^2 \right)\right|_{0}^2 \\ &=(-\tfrac12+2-\tfrac13)+(4-2)=4-\tfrac56 \\ &=\tfrac{19}6 . \end{align}

Another possible way is to split the area into three parts:

\begin{align} [ABCD]&=[\triangle AOD]+[\triangle BCD]+[ABO] \\ &=\tfrac12\,|AO|\cdot|OD|+\tfrac12\,|BD|\cdot|C_x-D_x| +\left|\int_{-1}^{0}f_1(x)\,dx\right| \\ &=\tfrac12\cdot1\cdot1+\tfrac12\cdot2\cdot2 +\left|\int_{-1}^{0}(x^2-1)\,dx\right| \\ &= \tfrac52 +\left|\left.(\tfrac13\,x^3-x)\right|_{-1}^{0}\right| \\ &= \tfrac52 +\left|\tfrac13-1\right| = \tfrac{5}2+\tfrac23 =\tfrac{15}6+\tfrac46 =\tfrac{19}6 . \end{align}