求由$y=x^2-1$、$y=2x-1$、$y=x+1$围成区域的面积，仍不理解解法

Hey there! Let's break this down step by step — I know figuring out enclosed areas with multiple curves can get confusing, so we'll take it slow and make sure every part makes sense.

Step 1: Find all intersection points #

First, we need to locate where each pair of curves cross each other, since these points mark the boundaries of our enclosed region.

• Intersection of $y=x^2-1$ and $y=2x-1$:
  Set the equations equal: $x^2 - 1 = 2x - 1$
  Simplify: $x^2 - 2x = 0 \implies x(x-2)=0$
  Solutions: $x=0$ (so $y=-1$) and $x=2$ (so $y=3$). Points: $(0, -1)$ and $(2, 3)$.

• Intersection of $y=x^2-1$ and $y=x+1$:
  Set equal: $x^2 -1 = x +1$
  Simplify: $x^2 -x -2=0 \implies (x-2)(x+1)=0$
  Solutions: $x=2$ (so $y=3$) and $x=-1$ (so $y=0$). Points: $(2, 3)$ and $(-1, 0)$.

• Intersection of $y=2x-1$ and $y=x+1$:
  Set equal: $2x -1 = x +1$
  Solution: $x=2$ (so $y=3$). Point: $(2, 3)$ (we already found this one!).

Our three distinct boundary points are: $(-1, 0)$, $(0, -1)$, and $(2, 3)$.

Step 2: Split the region into integrable intervals #

Looking at the curves, we split the enclosed area into two intervals based on our intersection points: $[-1, 0]$ and $[0, 2]$. We need to confirm which curve is the "upper boundary" and which is the "lower boundary" in each interval (this tells us what to subtract in our integrals).

For interval $[-1, 0]$: #

Pick a test point like $x=-0.5$:

• $y=x+1 = -0.5 +1 = 0.5$ (highest value)
• $y=x^2-1 = 0.25 -1 = -0.75$ (middle value)
• $y=2x-1 = -1 -1 = -2$ (lowest value)

Here, the upper boundary is $y=x+1$ and the lower boundary is $y=x^2-1$ (the line $y=2x-1$ sits below the parabola here and doesn't form the enclosed region's edge).

For interval $[0, 2]$: #

Pick a test point like $x=1$:

• $y=x+1 =1+1=2$ (highest value)
• $y=2x-1=2-1=1$ (middle value)
• $y=x^2-1=1-1=0$ (lowest value)

Here, the upper boundary is $y=x+1$ and the lower boundary is $y=2x-1$ (the parabola lies below the line here and doesn't form the enclosed region's edge).

Step 3: Calculate the integrals for each interval #

The area between two curves $y_{upper}$ and $y_{lower}$ from $a$ to $b$ is $\int_{a}^{b} (y_{upper} - y_{lower}) dx$.

Integral for $[-1, 0]$: #

$$
\begin{align*}
\int_{-1}^{0} [(x+1) - (x^2 -1)] dx &= \int_{-1}^{0} (-x^2 + x + 2) dx \
&= \left[ -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x \right]_{-1}^{0} \
&= 0 - \left( -\frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2(-1) \right) \
&= 0 - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) \
&= 0 - \left( -\frac{7}{6} \right) \
&= \frac{7}{6}
\end{align*}
$$

Integral for $[0, 2]$: #

$$
\begin{align*}
\int_{0}^{2} [(x+1) - (2x -1)] dx &= \int_{0}^{2} (-x + 2) dx \
&= \left[ -\frac{1}{2}x^2 + 2x \right]_{0}^{2} \
&= \left( -\frac{1}{2}(4) + 2(2) \right) - 0 \
&= (-2 + 4) \
&= 2
\end{align*}
$$

Step 4: Sum the areas to get the total enclosed area #

Add the results from both intervals:
$$
\text{Total Area} = \frac{7}{6} + 2 = \frac{7}{6} + \frac{12}{6} = \frac{19}{6} \approx 3.1667
$$

内容的提问来源于stack exchange，提问作者user527567