Maharashtra State Board · Class 12 · Mathematics · Part II Chapter 5

Application of Definite Integration — Formula Sheet

Board Formulas

13 formulas

1.Area Bounded by a Curve and the x-axis★
$A = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x, f (x) \geq 0 on [a, b]$
$A$ : Area of the region (sq. units) · $y = f (x)$ : Equation of the curve · $a, b$ : x-coordinates of the bounding vertical lines
Region between y = f(x), the x-axis and the lines x = a, x = b. Uses vertical strips of height y and width dx.
2.Area Bounded by a Curve and the y-axis
$A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y, g (y) \geq 0 on [c, d]$
$x = g (y)$ : Equation of the curve solved for x · $c, d$ : y-coordinates of the bounding horizontal lines
Region between x = g(y), the y-axis and the lines y = c, y = d. Uses horizontal strips; the limits are y-values.
3.Region Below the x-axis
$A = \int_{a}^{b} f (x) d x, f (x) \leq 0 on [a, b]$
The integral comes out negative when the curve lies below the axis; the area is its absolute value.
4.Curve Crossing the x-axis
$A = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$
Here f ≥ 0 on [a, c] and f ≤ 0 on [c, b]. Find every root c between a and b and split there; otherwise positive and negative parts cancel.
5.Area Between Two Curves (vertical strips)★
$A = \int_{a}^{b} [f (x) - g (x)] d x, f (x) \geq g (x) on [a, b]$
$f (x)$ : Upper curve · $g (x)$ : Lower curve
Always upper curve minus lower curve. The limits a and b are usually the x-coordinates of the points of intersection.
6.Area Between Two Curves (horizontal strips)
$A = \int_{c}^{d} [ϕ (y) - ψ (y)] d y, ϕ (y) \geq ψ (y) on [c, d]$
$ϕ (y)$ : Right-hand curve, x = φ(y) · $ψ (y)$ : Left-hand curve, x = ψ(y)
Right curve minus left curve, with y-limits. Useful when the curves are naturally given as x in terms of y.
7.Limits from Points of Intersection
$f (x) = g (x) \Rightarrow x = a, x = b$
Solve the two equations simultaneously. Check which curve is on top by substituting one x-value between a and b.
8.Quarter-Circle Integral
$\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}$
Comes from ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + c. Needed for every circle and ellipse question.
9.Area of a Circle★
$x^{2} + y^{2} = a^{2} : A = 4 \int_{0}^{a} a^{2} - x^{2} d x = π a^{2}$
The circle is symmetric about both axes, so take 4 × the first-quadrant area, where y = √(a² − x²).
10.Area of an Ellipse★
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x = π ab$
$a, b$ : Semi-axes along the x- and y-axes (semi-major and semi-minor when a > b)
In the first quadrant y = (b/a)√(a² − x²). For a = b it reduces to the circle πa².
11.Parabola Cut Off by its Latus Rectum
$y^{2} = 4 a x, x = a : A = 2 \int_{0}^{a} 2 a x d x = \frac{8 a ^{2}}{3}$
The parabola is symmetric about the x-axis, so double the area above it, where y = 2√(ax).
12.Parabola and a Line Through the Vertex
$y^{2} = 4 a x, y = m x (m > 0) : A = \int_{0}^{4 a / m^{2}} (2 a x - m x) d x = \frac{8 a ^{2}}{3 m ^{3}}$
The curves meet at (0, 0) and (4a/m², 4a/m). The parabola is the upper curve between them. Use it to check your answer, but show the integration in the exam.
13.Area Between y² = 4ax and x² = 4ay
$A = \int_{0}^{4 a} (2 a x - \frac{x ^{2}}{4 a}) d x = \frac{16 a ^{2}}{3}$
The parabolas meet at (0, 0) and (4a, 4a). y² = 4ax is the upper curve between them. More generally, y² = 4ax and x² = 4by enclose 16ab/3.