Finding the Area Between Curves Expressed as Functions of x
Introduced| English |
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| area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ |
The area trapped between two curves
- A single integral gives the area between a curve and the $x$-axis. What about the area between two curves?
- Slice the region into thin vertical strips; each strip's height is the gap between the curves.
- Add up (integrate) those heights across the interval — that's the area between two curves 曲线间面积.
- The recipe is "top minus bottom, integrated."
Top minus bottom
- If $f$ is the upper curve and $g$ the lower on $[a,b]$:
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$$A=\int_a^b \big(f(x)-g(x)\big)\,dx$$
- Each vertical strip has height $f(x)-g(x)$ and width $dx$.
- Always subtract upper minus lower, so the height (and the area) stays positive.
A line above a parabola
y = ax² + bx + c
Between the intersection points the line sits above the parabola — the vertical gap, integrated, is the enclosed area.
The area between an upper curve $f$ and lower curve $g$ on $[a,b]$ is...
Upper minus lower, integrated.
Find the limits from intersections
- The bounds $a$ and $b$ are usually where the curves cross — set $f(x)=g(x)$ and solve.
- Those intersection $x$-values are the left and right edges of the enclosed region.
- Between them, decide which curve is on top (test a point).
- Then integrate the top-minus-bottom difference over $[a,b]$.
Where do $y=x+2$ and $y=x^2$ intersect?
$x^2-x-2=0\Rightarrow x=-1,2$.
To decide which curve is on top between intersections, you...
Plug in a test point to see which is higher.
The limits of integration come from setting the two curves ____.
Solve $f(x)=g(x)$ for the intersection $x$-values.
It works even below the axis
- Because you take upper minus lower, the formula handles regions below the $x$-axis automatically.
- The $-g(x)$ term lifts everything, so no separate "signed area" bookkeeping is needed.
- A region entirely under the axis still gives a positive area with top-minus-bottom.
- That's the advantage of the gap-height approach.
The area between $y=x+2$ and $y=x^2$ is $\int_{-1}^2 (x+2-x^2)\,dx$. Compute it (a decimal).
The integral evaluates to $\tfrac92=4.5$.
Integrating lower minus upper gives the negative of the true area.
Reversing the order flips the sign.
Always subtract upper minus lower — if you reverse it, you get a negative of the true area. Check which curve is on top by testing an $x$-value between the intersections (they can swap, which is the next lesson). And the limits come from setting the curves equal, not from the $x$-axis.
Find the area between $y=x+2$ and $y=x^2$.
- Intersections: $x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow x=-1,2$.
- On $(-1,2)$ the line is on top (test $x=0$: line $=2>0=$ parabola).
- $A=\displaystyle\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\Big[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\Big]_{-1}^{2}=\tfrac{9}{2}$.
The area between two curves (functions of $x$) is $A=\int_a^b\big(f(x)-g(x)\big)\,dx$, integrating upper minus lower over vertical strips. Find $a,b$ where the curves intersect, decide which is on top, and integrate the gap. Top-minus-bottom keeps the area positive automatically.