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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 on [a, b] ATTAINS its average at some c in [a, b]
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
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
Unit 8 tools
Challenge bank
1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

0 mastered · 0 to revisit · 60 total
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Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 13 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from. AB students are served only the AB items; BC students get the whole unit.

20questions
13topics
15codes covered
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Course so far · AB

Check what stuck.

Units 1 through 8, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 45-question section, your choice. Even coverage means this is a retention check rather than a score estimate. Follows the track you chose on the course page.

20 or 45questions
81topics
116codes covered
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