The area is the definite integral of y = 2x from 0 to 3: Area = the integral from 0 to 3 of 2x dx = [x^2] from 0 to 3 = (3^2) - (0^2) = 9. So the required area is 9 square units. (This agrees with the area of the triangle of base 3 and height 6, which is (1/2)(3)(6) = 9.)

Area = integral(a to b) y dx

The equation x^2 + y^2 = 16 represents a circle centred at the origin with radius r = 4 (since r^2 = 16). The area enclosed by a circle of radius r is pi r^2. So the area = pi (4)^2 = 16 pi square units. (By integration, the area = 4 times the integral from 0 to 4 of sqrt(16 - x^2) dx = 16 pi.)

Area of circle = pi r^2

If a region is symmetric about the x-axis, the y-axis or the origin, we can compute the area of one symmetric part and then multiply by the appropriate factor to get the total area. For example, a circle x^2 + y^2 = a^2 is symmetric about both axes, so its total area equals 4 times the area of the part lying in the first quadrant. Using symmetry reduces the amount of integration needed and simplifies the calculation.

A differential equation is an equation that involves an independent variable, a dependent variable and the derivatives of the dependent variable with respect to the independent variable. The order of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the power (exponent) of the highest order derivative, provided the equation is a polynomial in the derivatives (free from radicals and fractions in the derivatives).

The highest order derivative present in the equation is d2y/dx2 (the second derivative), so the order of the differential equation is 2. The power (exponent) of this highest order derivative is 3, and the equation is a polynomial in the derivatives, so the degree is 3. Hence the order is 2 and the degree is 3.

The general solution of a differential equation is the solution that contains as many arbitrary constants as the order of the differential equation; it represents a whole family of curves. A particular solution is obtained from the general solution by giving definite values to these arbitrary constants using given initial or boundary conditions; it represents one specific member of the family.

Separate the variables: dy/(1 + y^2) = dx/(1 + x^2). Now integrate both sides: the integral of dy/(1 + y^2) = the integral of dx/(1 + x^2). This gives tan inverse (y) = tan inverse (x) + C, where C is the constant of integration. This is the general solution of the differential equation.

integral dy/(1 + y^2) = tan inverse (y)

Separate the variables by cross-multiplying: y dy = x dx. Integrate both sides: the integral of y dy = the integral of x dx, giving y^2/2 = x^2/2 + C1. Multiplying by 2: y^2 = x^2 + C (where C = 2 C1). So the general solution is y^2 - x^2 = C, which represents a family of hyperbolas.

A differential equation of the form dy/dx = f(x, y) is called homogeneous if the function f(x, y) can be written as a function of the ratio y/x only (that is, f depends on x and y only through y/x). To solve it, we use the substitution y = v x (so v = y/x), which gives dy/dx = v + x (dv/dx). This substitution reduces the equation to a variable separable form in v and x, which can then be integrated.

y = v x, dy/dx = v + x(dv/dx)

A linear differential equation of the first order and first degree has the standard form dy/dx + P y = Q, where P and Q are functions of x only (or constants). It is solved using the integrating factor (IF) = e^(integral of P dx). Multiplying the equation by the integrating factor makes the left side an exact derivative, and the solution is y x (IF) = integral of [Q x (IF)] dx + C.

IF = e^(integral of P dx)

Here P = 1/x and Q = x. The integrating factor IF = e^(integral of (1/x) dx) = e^(ln x) = x. Multiply through by x (or use the solution formula): y x (IF) = integral of [Q x (IF)] dx, that is, y x = integral of (x times x) dx = integral of x^2 dx = x^3/3 + C. Dividing by x gives the general solution y = x^2/3 + C/x.

y (IF) = integral of Q (IF) dx + C

A scalar quantity is one that has only magnitude and no direction (for example, mass, length, temperature). A vector quantity is one that has both magnitude and direction and obeys the laws of vector addition (for example, displacement, velocity, force). The position vector of a point P with coordinates (x, y, z) with respect to the origin O is the vector OP = x i + y j + z k, and its magnitude is sqrt(x^2 + y^2 + z^2).

|OP| = sqrt(x^2 + y^2 + z^2)

The magnitude of a = 2 i + 3 j + 6 k is |a| = sqrt(2^2 + 3^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7. A unit vector in the direction of a is obtained by dividing a by its magnitude: a hat = a/|a| = (1/7)(2 i + 3 j + 6 k) = (2/7) i + (3/7) j + (6/7) k.

a hat = a/|a|

A unit vector is a vector whose magnitude is 1; it is used to specify direction. A zero (null) vector is a vector whose magnitude is zero and whose direction is indeterminate; it is denoted by 0. Collinear vectors are two or more vectors that are parallel to the same line, irrespective of their magnitudes and directions; such vectors can be written as scalar multiples of one another.

The scalar (dot) product of two vectors is the sum of the products of their corresponding components: a . b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3. The dot product is a scalar quantity and also equals |a||b| cos theta, where theta is the angle between the vectors.

a . b = a1 b1 + a2 b2 + a3 b3

The angle theta satisfies cos theta = (a . b)/(|a| |b|). Here a . b = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1. Also |a| = sqrt(1 + 1 + 1) = sqrt(3) and |b| = sqrt(1 + 1 + 1) = sqrt(3). So cos theta = 1/(sqrt(3) x sqrt(3)) = 1/3. Therefore theta = cos inverse (1/3).

cos theta = (a . b)/(|a| |b|)

The cross product is found by expanding the determinant with rows i, j, k; then the components of a; then the components of b. a x b = i[(1)(1) - (1)(-1)] - j[(1)(1) - (1)(1)] + k[(1)(-1) - (1)(1)] = i(1 + 1) - j(1 - 1) + k(-1 - 1) = 2 i - 0 j - 2 k = 2 i - 2 k. Its magnitude is |a x b| = sqrt(2^2 + 0^2 + (-2)^2) = sqrt(8) = 2 sqrt(2).

a x b = determinant of (i, j, k; a1, a2, a3; b1, b2, b3)

The scalar projection (component) of vector a on vector b is given by (a . b)/|b|. It represents the length of the shadow of a along the direction of b. If the dot product a . b = 0 (and neither vector is zero), then cos theta = 0, so theta = 90 degrees; this means the two vectors are perpendicular (orthogonal) to each other.

projection of a on b = (a . b)/|b|

Two vectors are perpendicular if their dot product is zero. So a . b = (2)(1) + (lambda)(-2) + (1)(3) = 2 - 2 lambda + 3 = 5 - 2 lambda. Setting this equal to zero: 5 - 2 lambda = 0, so 2 lambda = 5 and lambda = 5/2.

a . b = 0 for perpendicular vectors

The direction cosines of a line are the cosines of the angles that the line makes with the positive directions of the x-, y- and z-axes; they are usually denoted by l, m and n. The direction ratios are any set of numbers proportional to the direction cosines. The direction cosines always satisfy the relation l^2 + m^2 + n^2 = 1.

l^2 + m^2 + n^2 = 1