CBSE · Class 12 · Mathematics · Chapter 8

Application of Integrals — Formula Sheet

Board Formulas

13 formulas

1.Elementary Strip
$d A = y d x (vertical strip), d A = x d y (horizontal strip)$
Think of the region as many thin strips of height y and width dx. Adding them up (integrating) gives the area.
2.Area Under a Curve (about 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 (square units) · $a, b$ : Left and right ordinates (x = a, x = b), with b > a · $f (x)$ : Curve y = f(x) bounding the region from above
Region bounded by y = f(x), the x-axis and the ordinates x = a and x = b. Use vertical strips.
3.Area About the y-axis
$A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y, g (y) \geq 0 on [c, d]$
$c, d$ : Lower and upper lines y = c and y = d, with d > c · $g (y)$ : Curve x = g(y) bounding the region on the right
Region bounded by x = g(y), the y-axis and the lines y = c and y = d. Use horizontal strips and write x in terms of y.
4.Curve Below the x-axis
$A = \int_{a}^{b} f (x) d x, f (x) \leq 0 on [a, b]$
Below the axis the definite integral comes out negative. Take its absolute value, because area is always positive.
5.Curve Crossing the x-axis★
$A = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x, f (c) = 0$
Split at every point where the curve crosses the x-axis, then add the absolute values. For a line y = mx + k, the crossing is at x = −k/m. Integrating straight from a to b gives the NET value, not the area.
6.Odd Function on a Symmetric Interval
$f (- x) = - f (x) : \int_{- a}^{a} f (x) d x = 0 but A = 2 \int_{0}^{a} f (x) d x$
The parts above and below the axis cancel in the integral but not in the area. The area formula assumes f keeps one sign on [0, a]. Example: y = x³ from −1 to 1 has area 1/2, not 0.
7.Area Under y = kxⁿ
$\int_{0}^{b} k x^{n} d x = \frac{k b ^{n + 1}}{n + 1}, n \geq 0$
A quick check for regions under y = x², y = x³, y = √x and similar curves, starting at x = 0. Here k > 0 and b > 0.
8.One Arch of the Sine Curve
$\int_{0}^{π} sin x d x = 2$
Each arch of y = sin x or y = cos x encloses 2 square units with the x-axis. So the area between y = sin x and the x-axis from 0 to 2π is 4, even though the integral is 0.
9.Standard Integral for Circle and Ellipse★
$\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$
Comes from Chapter 7. Every circle and ellipse area depends on it, so learn it exactly, including the a²/2 factor.
10.Quarter-circle Integral
$\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}$
At x = a the first term vanishes and sin⁻¹(1) = π/2. At x = 0 both terms are 0. This is the first-quadrant area of x² + y² = a².
11.Area of a Circle
$x^{2} + y^{2} = a^{2} : A = 4 \int_{0}^{a} a^{2} - x^{2} d x = π a^{2}$
$a$ : Radius of the circle (units)
The circle is symmetric about both axes. Take y = +√(a² − x²) in the first quadrant and multiply by 4.
12.Area of an Ellipse★
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = 4 \cdot \frac{b}{a} \int_{0}^{a} a^{2} - x^{2} d x = π ab$
$a$ : Semi-axis along x (units) · $b$ : Semi-axis along y (units)
Solve for y = (b/a)√(a² − x²) in the first quadrant. A circle is the special case a = b.
13.Parabola y² = 4ax up to x = h
$A = 2 \int_{0}^{h} 2 a x d x = \frac{8}{3} a h^{3/2}$
$a$ : Parabola parameter in y² = 4ax (a > 0) · $h$ : Ordinate x = h closing the region (h > 0)
Region bounded by y² = 4ax and the ordinate x = h. The parabola is symmetric about the x-axis, so double the upper half y = 2√(ax). For h = a (the latus rectum) the area is 8a²/3.