Average Value of a Function
How do you find the average height of a curve over an interval? You "smush" the area into a rectangle.
"Integral divided by Interval Width"
- Always show (b-a) as the width.
- Units: average value has the same units as f(x).
- Common FRQ follow-up: find c where f(c)=favg.
Find the average value of f(x) = x² on [0, 3].
- Interval Width: b - a = 3 - 0 = 3.
- Integrate: ∫03 x² dx = [x³ / 3]03 = 27/3 - 0 = 9.
- Divide: Average = 9 / 3 = 3.
Find a value c in [0,3] such that f(c)=favg.
- Set x² = 3.
- Solve: x = √3 (the only solution in the interval).
On FRQs: one equation + one value in the interval = full credit.
1) Find favg for f(x)=\sin x on [0, \pi].
2) For the same function, find c where f(c)=favg (give one value).
Show Answers
1) favg = \frac{1}{\pi-0}\int_0^\pi \sin x\,dx = \frac{1}{\pi}[ -\cos x ]_0^\pi = \frac{1}{\pi}(2)=\frac{2}{\pi}.
2) Solve \sin c = \frac{2}{\pi}. One solution is c=\arcsin\left(\frac{2}{\pi}\right) (in [0,\pi]).
Position, Velocity, Acceleration (Integrals)
In Unit 4, we differentiated position to get velocity. Now, we integrate velocity to get position.
| Concept | Formula | Meaning |
|---|---|---|
| Displacement | ∫ab v(t) dt | Net change in position. (Can be 0 if you return home). |
| Total Distance | ∫ab |v(t)| dt | Total ground covered. (Odometer reading). Always positive. |
| Current Position | s(t) = s(a) + ∫at v(x) dx | Start Position + Displacement. |
- Find where v(t)=0 (turning points) before doing total distance.
- Sign matters: negative velocity means moving backward (displacement can be negative).
- Units: if v is m/s and t is seconds, the integral is meters.
Let v(t)=t²-4 on 0 ≤ t ≤ 3.
- Where does it change sign? t²-4=0 → t=2.
- Displacement: ∫03(t²-4)dt = [t³/3 - 4t]03 = (9 - 12) = -3.
- Total Distance: split at t=2:
∫02(4-t²)dt + ∫23(t²-4)dt
Because t²-4 is negative on [0,2] and positive on [2,3].
1) If s(0)=5 and v(t)=2t, find s(3).
2) Write (don’t compute) an expression for total distance traveled if v(t)=\sin t on [0, \pi].
Show Answers
1) s(3)=5+∫032t\,dt = 5+[t²]03=14.
2) ∫0\pi|\sin t|\,dt (here it’s all positive, so =∫_0^\pi \sin t\,dt).
Area Between Curves
To find the area between two functions, we slice it into rectangles. Height = Top - Bottom.
Vertical Slices (dx)
Top Function minus Bottom Function
Horizontal Slices (dy)
Right Function minus Left Function
- Step 0: find intersection points (solve f(x)=g(x)) to get limits.
- Pick the easier variable: if one way needs splitting into 2+ integrals, try the other.
- Area is positive: use Top-Bottom (or Right-Left) so you don’t get a negative answer.
- Intersections: x = x² → x(x-1)=0 → x=0,1.
- Top vs Bottom on [0,1]: x is above x².
- Area: ∫01(x - x²)dx.
AP loves “set up the integral” questions. This is already full credit for setup.
Region bounded by x=y² and x=2y.
- Intersections: y² = 2y → y(y-2)=0 → y=0,2.
- Right - Left: right is x=2y, left is x=y².
- Area: ∫02(2y - y²)dy.
1) Set up (don’t compute) the area between y=\ln x and y=1 on [1,e].
2) If you switch to dy, what would you need first?
Show Answers
1) On [1,e], top is 1 and bottom is \ln x, so Area = ∫1e(1-\ln x)\,dx.
2) You’d need x as functions of y (solve y=\ln x → x=e^y) and determine right/left limits in y.
Volumes with Cross Sections
Imagine a 3D solid built on a base region R. Volume is the integral of area of each slice.
- Sketch the base region and label Top(x) and Bottom(x).
- Write the slice width: s = Top(x) - Bottom(x).
- Convert s into an area formula (square, triangle, semicircle, etc.).
- Integrate Area(x) over the correct limits.
If the base of the slice is s = Top(x) - Bot(x):
- Square: A = s²
- Semicircle: diameter = s → A = (π / 8)s²
- Equilateral Triangle: side = s → A = (√3 / 4)s²
- Isosceles Right Triangle: legs = s → A = (1/2)s²
- Rectangle: A = s · h (height usually given).
Base region is between y=x (top) and y=0 (bottom) from x=0 to x=2. Cross sections perpendicular to the x-axis are squares.
- Slice width: s = Top - Bottom = x - 0 = x.
- Area of square: A(x)=s²=x².
- Volume: V=∫02x²\,dx.
On FRQs, labeling s clearly is often the difference between 3/4 and 4/4.
1) Base is between y=4-x² and y=0. Cross sections are semicircles with diameter in the base. Set up V.
Show Answer
Intersections: 4-x²=0 → x=±2. Diameter s= (4-x²)-0. Semicircle area: A=(π/8)s².
V = ∫-22 (π/8)(4-x²)² \, dx.
Rotational Volume: The Disc Method
Use disc method when the region is flush against the axis of rotation (no hole).
Think: sum of circle areas (πR²).
- Radius is a distance to the axis. If rotating around y=k, radius becomes |f(x)-k|.
- Write: “Radius = (top curve) − (axis)” (or “axis − curve”) so graders see the geometry.
Region under y=x from x=0 to x=2 rotated about the x-axis.
- Radius: R(x)=x.
- Volume: V=π∫02x²dx.
1) Set up the volume when the region under y=\sqrt{x} from 0 to 4 rotates about the x-axis.
Show Answer
R(x)=\sqrt{x} so V=π∫04(\sqrt{x})²dx = π∫04x\,dx.
Rotational Volume: The Washer Method
Use washer method when there is a gap between the region and the axis. The slice looks like a donut.
"Big Radius Squared minus Little Radius Squared"
Distance from Outer Curve to Axis.
Distance from Inner Curve to Axis.
Warning: Calculate R² - r². Do NOT calculate (R - r)². That is Algebra murder.
- Write radii as distances: R(x)=|top-axis|, r(x)=|bottom-axis| (depending on picture).
- If rotating about y=k, your radii become |f(x)-k| and |g(x)-k|.
Region between y=2 (top) and y=x (bottom) from x=0 to x=2, rotated about the x-axis.
- Outer radius: R(x)=2.
- Inner radius: r(x)=x.
- Volume: V=π∫02(2² - x²)\,dx.
1) Set up the volume when the region between y=x² and y=4 rotates about the x-axis.
Show Answer
Intersections: x²=4 → x=±2. Outer radius R=4, inner radius r=x².
V=π∫-22(4²-(x²)²)\,dx = π∫-22(16-x⁴)\,dx.
Arc Length BC Only
Arc length is the physical length of a curve. We approximate with tiny hypotenuses and integrate.
You MUST take the derivative first, then plug into the formula.
- Most arc length integrals are ugly. On AP, you may only need the setup, or you may be told to approximate.
- Perfect-square check: if 1+(f')² becomes a perfect square, the integral gets much easier.
- Calculator note: If asked to “approximate,” use a calculator and show the integral you evaluated.
- Derivative: f'(x)=3x².
- Arc length: L=∫01√(1+(3x²)²)\,dx = ∫01√(1+9x⁴)\,dx.
This is exactly the kind of FRQ where they give full credit for correct setup and may ask for a numerical approximation.
1) Set up (don’t compute) the arc length of y=\ln x on [1,e].
Show Answer
f'(x)=1/x, so L=∫1e√(1+(1/x)²)\,dx.