Key takeaway: The area between curves from x = m to x = n is found by integrating the upper function minus the lower function over that interval, so the result is never negative.
Grade 12 – AP Calculus ABAP Calculus AB Unit 8Calculus
Definition of Area Between Curves
The area between curves measures the region enclosed by two graphs. Slice the region into thin vertical strips: each strip has height equal to the top curve minus the bottom curve, and width \(dx\). Adding the strips is a definite integral.
The limits usually come from the points where the two graphs meet, found by setting the functions equal. If the curves cross inside the interval, the region must be split at each crossing so the top and bottom roles are kept straight.
Formula and notation
When \(T(x)\ge B(x)\) on \([m,n]\):
\[ \text{Area}=\int_{m}^{n}\big[T(x)-B(x)\big]\,dx. \]
Here \(T\) is the top curve and \(B\) the bottom one. Subtract in that order, so every strip height is positive.
Key facts
- Find the limits by solving T(x) = B(x) when the region is bounded by the curves.
- Sketch first to decide which function is on top over each interval.
- If the curves cross, split the integral at the crossing and keep top minus bottom each time.
- Horizontal strips (right minus left) are easier when curves are given as x in terms of y.
Area Between Curves example
Find the area enclosed by the line \(y=x+6\) and the parabola \(y=x^2\).
-
Where do the graphs meet? Setting the heights equal and factoring gives \((x-3)(x+2)=0\), so the limits are \(-2\) and \(3\).
-
On \([-2,3]\) the line is on top: test \(x=0\), where the line gives 6 and the parabola gives 0.
-
Integrate: \(\int_{-2}^{3}(x+6-x^2)\,dx=[\frac{x^2}{2}+6x-\frac{x^3}{3}]_{-2}^{3}\).
-
Upper limit: \(4.5+18-9=13.5\). Lower limit: \(2-12+\frac{8}{3}=-\frac{22}{3}\). Difference: \(13.5+\frac{22}{3}=\frac{125}{6}\).
The enclosed area is \(\frac{125}{6}\approx 20.83\) square units.
Common mistake
Watch out
A student subtracts in the wrong order, integrating \(x^2-(x+6)\) from \(-2\) to \(3\), and obtains \(-\frac{125}{6}\).
Why it is wrong
An area cannot be negative. The negative sign shows that the bottom function was placed first. Always take top minus bottom, deciding which is which from a sketch or a test value, and the area between curves comes out positive.
Related terms and lessons
Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus