Applications of Integration
Thirteen topics, one idea: integrate a cross-section. The average value of a function is an integral divided by the interval; position needs the starting value added and total distance needs the absolute value; the area between curves is top minus bottom, with a split at every crossing; a cross section contributes its own area, so a square gives s² and a washer gives R² − r², never (R − r)²; and a radius is always measured from the axis you are revolving about.
AB exam 10-15%BC exam 5-10%13 topics
Topics
Key forms For every problem in this unit
Average value
the integral DIVIDED by b − a. A value, not an integral
Divide by what
b − a, the LENGTH: never b, never the sample count
MVT for integrals
a continuous f ATTAINS its average somewhere inside
Net change
the integral of a RATE is the CHANGE, not the amount
Final amount
starting value PLUS the integral. The constant is not optional
Displacement
integrate v itself; sign and all
Total distance
integrate |v|, so SPLIT wherever v = 0
Speed
|v|. It rises when v and a share a sign
Units
rate units TIMES the units of the variable integrated
Saying what it means
name the quantity, the interval, and the units
With respect to x
TOP minus BOTTOM, dx, limits are x-values
With respect to y
RIGHT minus LEFT, dy, limits are y-values
Which one
pick the variable that needs FEWER splits
Going to dy
solve for x FIRST; y-limits, not the old x-limits
Which is on top
TEST a point between the crossings. Do not eyeball it
Negative answer
the order was reversed. Area is never negative
The limits
solve f = g for ALL roots; given numbers are not bounds
Curves that cross
one integral per piece, each ordered on its own
Why splitting matters
one integral across a crossing CANCELS, it does not add
Absolute value
|f − g| says split; it is not a repair applied at the end
Every volume
integrate ONE SLICE'S AREA along the perpendicular axis
The side s
the DISTANCE between the boundaries: top minus bottom
Square section
s²
Rectangle
s times the stated multiple of s
Isosceles right △
s²/2 with s a LEG; s²/4 with s the hypotenuse
Equilateral △
(√3/4)s², not (√3/2)s²
Semicircle
(π/2)r² with r = s/2, so (π/8)s²
Disc
π times radius squared, integrated
Washer
π(R² − r²). NEVER π(R − r)²
Radius
a DISTANCE from the axis, so it is never negative
About y = k
|f(x) − k|, adjusted on BOTH radii
About x = k
solve for x, integrate dy, radius |g(y) − k|
Arc length (BC)
√(1 + (f′)²) dx. The 1, the square, the root
Distance travelled (BC)
the same integral: arc length of the path