Application of Definite Integration — Class 12 Maharashtra HSC Mathematics revision notes
A 3-page visual revision sheet for Application of Definite Integration, written for Class 12 Maharashtra HSC Mathematics. Formulas, diagrams and the traps that cost marks, laid out to be read the night before rather than the term before. Free, and it opens inside CapCompass.
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Application of Definite Integration in one look
the definite integral read as area, one boundary curve at a time, then two curves subtracted
Area Under a Curve
y=f(x) continuous, f(x)≥0 on [a,b]: the bounded area is exactly ∫y dx from a to b
strip picture: thin vertical strips, width dx, height y; sum (integrate) from a to b
same idea sideways for x=g(y): thin HORIZONTAL strips, width dy, length x
Area with vertical strips (dx) and horizontal strips (dy)
Strip
Region
Formula
dx, vertical
y=f(x), X-axis, x=a to x=b, f(x)≥0
A=∫_a^b y dx
dy, horizontal
x=g(y), Y-axis, y=c to y=d, g(y)≥0
A=∫_c^d x dy
Area under a curve, dx strips
Q. Find the area bounded by the curve y=x^2, the X-axis, and the lines x=0 and x=3.
A = ∫_0^3 x^2 dx
antiderivative: x^3/3
apply the limits: 3^3/3 - 0^3/3 = 27/3
Answer: 9 square units
Area under a curve, dy strips
Q. Find the area bounded by the curve x=y^2, the Y-axis, and the lines y=1 and y=2.
A = ∫_1^2 x dy = ∫_1^2 y^2 dy
antiderivative: y^3/3
apply the limits: 2^3/3 - 1^3/3 = 8/3 - 1/3
Answer: 7/3 square units
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What this Application of Definite Integration sheet covers
Area Under a Curve
Area with vertical strips (dx) and horizontal strips (dy)
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