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cos²x + sin²x= 1
When two or more terms in a binomial expansions are known and a relationship between them is given (in this case they are equal), in the expansion of (x + y)^n, where n is not given: to find n, have the terms, to each other: equating (in other situations one term may be two or else times bigger than the other(s)). Rearrange, deduce and calculate to find the value of n.
When integrating to find the area between a curve and an axis: areas under the x-axis will come out negative and areas above the x-axis will be positive. If all the area is under the x-axis, if the answer that you get for the area is -n, put the area as n. Sometimes, a region contains a part below the axis and a part above the axis; to deduce the part(s) which are above and the parts below the x-axis: find the roots. If a region (R) has a part above the x-axis (a) and a part below (b), the two parts should be treated (integrated) seperately. a = 4, b = -2: R = 6.
u1 + u2 + u3 + . . . + un (characters after 'u' subscripted) is written in sigma notation with a sigma, below it: r= 1, on top of it: n, and in front of it ur (r ss).
A sigma has k = 1 at its buttom, n at its top and c in front of it, the sum is nc.
{cΣ(i = n) ^k fi (i ss) = Σ(i = n) ^k cfi (i ss).
{Σ(i=0)^k ai (i ss) + Σ(i=0)^k bi (i ss) = Σ(i=0)^k ai + bi (i ss).
A differential equation is an equation that contains a derivative.
To verify that something solves an equation, you need to substitute it into the equation and show that you get 0 (or other value if appropriate).
To solve or to find the general equation from a differential equation: integrate.
If y=f(x) and f(0) = 0, then y = 0 when x = 0.
To find the set of values on a graph at which the gradient is negative, solve dy/dx < 0 and to find the set of values at which the gradient is positive, solve dy/dx > 0, to in general get a or a set of interval(s).
A geometric progression is a sequence where each term is and the following term are in the ration 1: r; r is known as the common ratio of the sequence. The nth term of a geometric progression, where a is the first term and r is the common ratio, is: ar^(n-1).
The sum of the first n terms of a geometric progression is: a(1 - r^n )/ 1 – r.
The sum to infinity of a geometric progression: In geometric progressions where -1<r<1, thus where |r| < 1, the sum of the sequence as n tends to [] approaches a value infinity--. The answer to the sum to infinity of such a geometric progression is a/(1-r), that is the value that the series c[] to converges--.
The terms of a geometric series may tend to plus or minus infinity, 0 or none of those depending on the values of r and a.
The terms of an arithmetic series tend to plus or minus infinity(the series does not tend to a limit).
To find the limit to a sequence, for which the formula equates to un+1=f(un) (n,n+1 ss), replace un (n ss) in the equation with L and have the equation equating to L. Then solve for L.
Solving Multiple Angle Trigonometric Equations: if sin x° = b, other values that have a sine of b are sin(-180[]x), sin(0[]x) and sin(180[]x) sin(-180-x), sin(0+x), sin(180-x). There is a pattern: sin x°=sin(a±x), where a can be taken to start at 0 and increments both positively and negatively by 180; the numerative operation done at 0 is addition (+), and the operation [] between + and - alternates--. If you are given an 'x' value greater than 180, you may simplify it to, for example, 180-x or 540-x etc. (e.g sin 210° could be changed to sin -30° ).
The square of sine(x) is sine²x, (cos(x))² is cos²x, (tan(x))² is tan²x.
In the equation xloga = logb, x may be calculated using rearrangement to give x= loga/logb. For some situations, 'x' may appear on both sides, to find 'x': factorise. At times, it may be useful to use the rule 1/a^n = a^-n.
When taking the log of sides: log the non-scripted numbers and make like non-scripted numbers the term(s) that are exponents. The logged values and the modified previous exponents are multiplied.
For the curve y=ax²+bx+c the number of roots depends on the discriminant b... ²-4ac [b² - 4ac]= Δ.
For the curve y=ax²+bx+c the number of roots depends on the discriminant Δ = b² - 4ac. There are no roots if Δ < 0, there is one root if Δ = 0 and there are two roots if Δ > 0.
Where using discriminant for finding roots: a question may be to find the values of e.g k, for which the curve: e.g x² + kx + 2 has 0/1/2 roots. To answer: solve for k, Δ </=/> 0.
The exponential function, whose equation is equal to its derivative is written either exp(x) or e^x. y = e^x = dy/dx. Where k is a "constant": if y = e^kx: dy/dx = ke^(kx).
The inverse function of e^x is logex (e ss), often written ln x. The graphs of lnx and e^x are reflections of one another in the line f(x) = x.
Logarithms are another way of writing indices. To take the log of m = ab^t would give log m = t x log b + log a.
If a = b^c, then c = logba (b ss); meaning that the logarithm of a with respect to base b is c.
Finding the answer to log2(ss)7, may be done using the formula: logcb/logca (c ss) = loga(ss)b. The answer is logc(ss)7/logc(ss)2 (they may ask you to convert to a base, but otherwise you may pick one). log10(ss)7/log10(ss)2 may be written as log 7/log 2; loge(ss)7/loge(ss)2 may be written as ln 7/ ln 2.
The logarithm laws: (a ss) logax + logay = loga(xy); logax – logay = loga(x/y); loga(x^n) = nlogax; loga(1/y) = -loga(y). The base (a) may be ignored ( though it may be best if it's written), for with these laws of logarithms, it doesn't matter what the base is, as long as all of the logs are to the same base.
Logarithm laws (continued): logn(x)/logn(m) (n ss) = logmx (m ss).
With changing the base: let a, c, and b be p... ositive numbers that are r... eal such that neither a nor c equals 1; b is greater than 0.
With some logarithm equations, it may be better to state that a term (eg log2(ss)x in log2(ss)x + 1/log2(ss)x = 2) is equal to a variable.
With binomial expansions, a question may ask for the expansion of brackets up to the term in x^a (where a is a number). To do this expand each bracket using the binomial theorem, ignoring terms with higher powers than a, then having the brackets multiplied, again ignoring higher powers than a.
Finding the possible factors of a polynomial: if a polynomial (p(x)) has a factor: ax+b, then a factor of the highest power of the leading coefficient coefficient is a and b is a factor of the constant. If ax+b is a factor then a root is -b/a, thus p(-b/a) would equal 0.
The natural base is e; loge(ss)x = ln x.
All exponential curves (y=ak^x) have the property of passing through the point (0,a) whatever the value of k.
The sine graph: domain: (∞,-∞), range: [-1,1].
The cos graph: domain: (∞,-∞), range: [-1,1].
The tan graph: domain: all values except the ones that aren't real or odd multiples of π/2, range: [-∞,∞].
Arithmetic sequences may be defined iteratively, so that each term may be calculated from the previous term, or in closed form, such that there is a formula for the nth term.
The formula for the sum of the first n terms is Sn(ss) = n/2(2a+(n-1)d).
The amount of terms less than y, may be calculated with the inequality a+(n-1)d < y (with a and d turned to numerical values). If the calculated values aren't integers, round down. This technique may be used for other inequalities.
Instead of sigma, the symbol used maybe S.
The series of a sequence, from the first term to the nth terms (where u1[ss] is the first term, written in summation notation is Σr=1^n ur(ss).
When factorising cubic equations where one root (-o) is known: the cubic equation is f(x); f(x) = ax^3 + bx² + cx + d = (x+o)(px²+qx+r). The value of "p" is equal to the value of "a", the value of "q" may be deduced with the equation b = op+q, the value of "r" may be deduced with the formula: c = oq+r.
If (1/a)^n < b: n > log b / log a. The sign changed as there was division by log(1/a) which is negative.
When the polynomial expression p(x) is divided by (ax+b), if p(-b/a)=0, then ax+b is a factor. If p(x) is divided by ax+b with a remainder of n: p(-b/a) = n.
The polynomial p(x) = 3x^3 + ax² - bx + 1. If the remainder when p(x) is divided by x-2 is 3 and the remainder when p(x) is divided by x+1 is 4, find a and b. As p(2) = 3: 24 + 4a -2b +1 = 3: 4a - 2b = -22. As p(-1) = 4: -3+a+b+1=4: a + b = 0. 4a-2b=-22 (1) a+b=0 (2); to find a and b, solve the simultaneous equations. Then write the value of p(x).
When a polynomial f(x) is divided by x-a, the remainder is f(a).
In p(x) = q(x)s(x) + r, the quotients can be both q(x) (with division by s(x)) or s(x) (with division by q(x)), and the remainder is r.
A graph has the equation y = 2/x-3, the number of transformations from y=1/x amount to 2, a translation in the x direction by 3, and a scaling in the y direction by 2. (Possibly) the vertical asymptote is x=3.
"Often we can simplify a logarithmic expression so that it becomes either a single log, or a constant plus a log."
If log2(ss)3 = p: express log2(ss)216 in terms of p; log2(ss)216 = log2(ss)8 + log2(ss)27; as n log a = log a^n, both terms may be simplified with the same multiplier to give 3log2(ss)2 + 3log2(ss)3 = 3 + 3p, you may rearrange from there.
(2 ss) Simplify log2(a/b). log2(a/b) = (with subtraction) log2(ss)a - log2(ss)b.
Some exponential equations can be factorised in linear factors. The simplest can be factorised into quadratic equations. We then put each factor equal to zero and solve it. Example solve e^2x - 9e^x + 20 = 0; factorise to get (e^x - 4)(e^x - 5). If e^x = 4: x= ln 4, if e^x = 5: x = ln 5.
Some exponential equations can be factorised in linear factors, put each factor equal to zero and solve it. It may be convenient to have, where a^nx is the leading term, a substitution where p(or another variable) = a^ (nx/2) (or other if appropriate). Example solve 3e^4x - 8e^2x -3 = 0. Substitute p = e^2x to get 0 = 3p²-8p-3 and factorise to have (3p+1)(p-3); p = -1/3 or 3. If e^2x = -1/3: x= 0.5ln(-1/3) and if e^2x = 3: x = 0.5ln3.
To solve cos2x = sine (x), use one of the quadratic trigonometric formulae and substitute to turn the equation into sine (x)= 1-2sine²x, rearrange to get 0= 2sine²x+sine(x)-1, which factorises to (2sine(x)-1)(sine(x)+1).
Bearings may involve using t... rigonometry, generally both or either of the [] or [] rules sine and cosine rules.
More complicated questions may involve simultaneous equations: x^3-ax^2+bx+1: find the value of a and the value of b and factorise f(x), which has 1 and -3 as roots. To solve replace x in f(x) with both 1 and -3, then to get the value of a and the value of b: solve the simultaneous equations. The polynomial f(x) is a cubic and has the two given roots so must have a third linear factor cx+d; by considering the coefficient of x^3: c= 1; by considering the value of the multiplication of 1 and -3 and the value of the number at the end of f(x) being 1, the value of d is -1/3.
A quantity is said to grow or decay exponentially if the quantity at the start of each time period is multiplied, to obtain the quantity at the end of the period or the start of the next period, by a c... constant factor. If the constant factor is less than 1, then the quantity is exponentially decaying. If the constant factor is greater than 1, the quantity is exponentially growing.
{We can estimate an integral using the [] or [] rule trapezium or the mid – ordinate rule. Sometimes however it is desirable to find underestimates and overestimates for an integral. The closer, the more accurately can the value be found and we may quantify, for the estimate, the error. An underestimate may be made for an integration between the x values of n and a: find the value of y that is least in each interval (the first three being [n,n+1]{n+1,n+2][n+2,n+3])
{In general, the kth summand in a binomial expansion may be defined as C(n, k − 1) a^(n − (k − 1))b^(k − 1). The coefficient is not C(n, k) but C(n, k − 1) as the first number in each row of Pascal's Triangle is not entry 1, but entry 0.
The binomial expansion can in certain circumstances give highly accurate estimates of certain powers or roots. This can happen for example when we want to find not too high an indice of a number that is close to an integer.
The absolute value of a number is how far away it is from 0; |-2| = 2.
d/dx is the differential of a function with respect to x, d/dy is the differential of a function with respect to y.
Example: 4^6x - 5*4^3x -4 = 0; let y= 4^3x. If y=4^3x in 4^6x - 5*4^3x -4: to find y solve the equation: y²-5y-14; y=7,-2. If 4^3x=7: 3x= log4(ss)7; x= (log4(ss)7)/3; if 4^3x=-2: x= (log4(ss)-2)/3. As there is no value of log4(ss)-2: the solution is x= (log4(ss)7)/3.
In general, for the integral of concave functions, the trapezium method gives an underestimate, while it gives an overestimate for convex functions.
Plotting lines from exponential equations: "The most convenient form in which to analyse the relationship between two variables x and y is the linear form, and we must see if, given a suspected relationship between two variables, we can construct a linear graph from the suspected relationship." Eg the relationship between mass(m) and time(t) is m=ab^t. To make the graph linear, of both sides, take the log. to get log m=t*log b + log a With the data previously given, if not, it may be best to present it in a form that's tabular. As the log of both sides have been takes, have the table recalculated. Using the data in the table, the value of log a and lob b can be deduced, giving the line's equation. If log a = p: a = 10^p, if log b = q: b = 10^q. As a = 10^p and b = 10^q log m = t*log b + log a: m = 10^(t*10^q + 10^p).
{Proof of the sine rule: the sine rule states that for a triangle (labelled like a7 ): sine A/a = sine B/b = sine C/c. "To prove it, start by drawing a perpendicular line from a vertex to the opposite side, say from B to b. Label this perpendicular "x". Then x is equal to both c sine A and a sine C. Thus c sine A = a sine C and sine A/a = sine C/c. Draw another line from a vertex to the opposite side: y, thus y is equal to both c sine B and b Sine c. Thus sine B/b = sine C/c, and as sine A/a = sine B/b: sine A/a = sine B/b = sine C/c. A8 .
Example: The first three terms of a geometric sequence are k+1, 3k-2 and 4k+4. Find k, the first term and the common ratio. The common ration is the value when the terms are divided by the terms before. [First terms: k+1, 3k-2, 4k+4] Hence the common ratio may be deduced with the equation: 3k-2/k+1 = 4k/3k-2. K may be deduced through rearrangement. The value k = 0 or k = 4; If k=4: the first three terms are 5, 10, 20. The first term is 5 and the common ratio is 2; if k=0: the first three terms are 1,-2, 4. The first term is 1 and the common ratio is -2. Example: The 1st, 2nd and 4th terms of a geometric sequence are the 1st, 2nd and 3rd terms of an arithmetic sequence. Find the common ratio of the geometric sequence. The 1st, 2nd and 4th terms of a geometric sequence may be written a, ar, ar^3. The difference of the sequence is equal to both ar-a and ar^3-ar, thus d=ar-a=ar^3-ar. The value of r may be deduced from ar-a=ar^3-ar through rearrangement; r=-1±√5/2, but if r<0, then the signs would alternate and it wouldn't be arithmetic; so r(the common ratio)=1+√5/2.
Typically we have to solve equations of the form: e^4x-7e^x+10=0. If we substitute e^2x = y; we obtain the normal quadratic equation: y²-7y+10, which factorises to (y-2)(y-5), so y=2,5. If y=2: e^2x=2; 2x= ln(2); x=ln(2)/2. If y=5: e^2x=5; x= ln(5)/2.
Example: 4^6x - 5*4^3x -4 = 0, have y= 4^3x. If y=4^3x in 4^6x - 5*4^3x -4: to find y solve the equation: y²-5y-14; y=7,-2. If 4^3x=7: 3x= log4(ss)7; x= (log4(ss)7)/3; if 4^3x=-2: x= (log4(ss)-2)/3. As there is no value of log4(ss)-2: the solution is x= (log4(ss)7)/3.
The sum of the first n terms of a geometric progression is: a(1 - r^n )/ 1 – r. If r is between 1 and -1, thus |r| < 1, then we may sum an infinite number of terms and obtain a proper answer. Since in the expression for Sn(ss), for |r| < 1: r^∞ = 0. If r to the power of infinity is 0, and Sn(ss) = a(1 - r^n )/1–r, then S∞(ss) = a/1-r.
The first term in a geometric sequence is 4 and the 4th term is 0.0625, find the least value of n, such that the difference between Sn(ss) and S∞(ss) is less than 10^-6. 0.0625 = 1/16 = 4r^ 3; r = 1/4. 10^-6 > |S ∞(ss) - Sn(ss)|. |S∞(ss) - Sn(ss)| = |a /(1-r) - a(1-r^n)/(1-r)| = (through simplification) |ar^n/1-r|. As "r" = 1/4, "a" = 4 and |ar^n/1-r| < 10^-6: through substitution |ar^n/1-r| may be changed to 4(1/4)^n/1 - 1/4 = 4(1/4)^n / (3/4). Rearranging gives (1/4)^n < 3/16 * 10^-6. You may rearrange for n by taking the logs to get n > log(3/16 * 10^-6) / log(1/4) > 11.17, but as n is an integer, n = 12.
To solve cos2x = sine (x), use one of the quadratic trigonometric formulae(cos2x=1-2sine²x) and substitute to turn the equation into sine (x)= 1-2sine²x, rearrange to get 0= 2sine²x+sine(x)-1, which factorises to (2sine(x)-1)(sine(x)+1). The value of x is either sine^-1(0.5) or sine^-1(-1). If an equation may not be factorised, it may be solved using the quadratic formula. Example: solve -.3cos2x=2sine(x); by using one of the formulae (cos2x=1-2sine²x) there may be the obtaining of the equation: 0= 2sine²x+2sinex-1; this may be solved by using the quadratic formula, one way through substitution of sinex with a variable. If p = sine x: 0 = 2p²+2p-1.
Solve sine2x=cosx. The equation sine2x=cosx leads to (through use of trigonometric identities) 0= 2sinxcosx-cosx, which factorises to give 0 = cosx(2sinex-1). There may be some trigonometric equations that are not actually quadratic, but still require manipulation: 3cosx=7sinx, to solve divide both sides by 7sinex to get 3[...] = [] /7 tanx = 1: x= tan^-1(7/3).
{PFT(1): Let p(x) be a polynomial of degree n; (x-a) is a factor of p(x) if and only if (x-a) divides p(x) or is, of p(x), a factor. Let p(x) be a polynomial and let a be a number.
{If (x-a) divides p(x), then the remainder on division of p(x) by (x-a) is 0 and there is q(x) which is a polynomial such that p(x) = (x-a)q(x), so that p(x)= 0 and (x-a), of p(x), is a factor. Now assume that x=a is a root of p(x), so that p(x) = 0.
{Perform on p(x) long division by (x-a) to obtain quotient q(x) and remainder r(x) to have p(x)= (x-a)q(x)+r(x). The degree of r(x) is less than the degree of (x-a)[?] so r(x) is a constant. Write r(x)= c.
{Substitute x=a into p(x) = (x-a)q(x)+r(x) to get p(a) = 0 (the value of c is thus 0). PRT: f(x) = (x-a)·q(x) + r(x), but as r(x) is a constant substitute it for c, so f(x)= (x-a)q(x)+r(x); with x being equal c: f(c)= r.
{The cosine rule states that a²=b²+c²-2bccos(A), to prove this draw (in a7 ) a line from B to b (y). Split b into half, to the left: label x, and to the right: label b-x. Pythagoras' theorem gives y² equals both c²-x² and a²-(b-x)²: c²-x² = a²-b²-x²+2bx, hence a² = b²+c²-2bx. As x =[]cos[] c cos A: a² = b²+c²-2bccosA.
We can estimate an integral using the [] or [] rule least in each interval (the first three being [n,n+1]{n+1,n+2][n+2,n+3]) and add up the areas of the rectangles with these heights. If the value of the sum is equal to o and i is the actual value of the integral: n <= i. A similar method may be used when finding overestimates, but in the intervals find the maximum value of y and adding the areas, with these heights, of the triangles. If the value of the sum is p: the the value of i may be written with inequalities as p>=i=>o.
If ab = c and on one axis of a graph was a while on the other was b: the area under the graph represents c (e.g the area under a speed/time graph is the distance travelled).
With the sine rule: when teh angle that you want is larger than the given angle, there are [] possible results 2--. That is because you can draw two possible [] with the data triangles--. In general, sin(180-x)° = sin x°. A triangle has vertices (in clockwised order) A, B and C (starting at the buttom left). BCA will be larger then BAC if []>[] BA>BC, and in such a situation there will be two possible results. Imagine putting a pair of compasses at B, then drawing an arc with centre B and radius BC. It wuold intersect AC at two points - C1 and C2 showing that there are two triangles (BAC1 and BAC2) in which a, c and A have the same value in both triangles. (That would not happen if BC>BA). If one value of C is y°, then another possible answer is 180°-y°.
If cosA=b, where b is a negative number, then A is (in terms of types of angles) an obtuse angle.
Where a can be taken to start at 0 and increments by 360: cos x° = cos(a±x). Note: the negative values of x have the same value as their positive counterparts by symmetry, thus cos(a±x)= cos(-a±x).
tan x° = tan (a+x°), where a can be taken to start at 0, and increments (both positively and negatively) by 180.
The amount in degrees that it takes for a wave to repeat its path is the wavelength. The amplitude of a curve is half of its total height.
With y=sin(bx+p) or y=cos(bx+p): the wavelength is affected by b and p affects the shift.
1^c or 1rad is 1 radian. One radian is defined by the angle at the centre of an arc, where the arc length is the same as the radius. Thus 1^c = []/[] 360°/2π, hence 2π^c = 360° and π ^c = 180°, therefore 1° = π^c/180.
If the radius is r and θ is given in radians: arc length = rθ (as θ/2π x 2πr = rθ). If θ is given in radians, the area of a sector = r²θ/2 (= θ/2π x πr²). (Make sure that the calculator is in radians mode).
In a triangle with hypothenuse 1, horizontal leg - x, and vertical leg - y: sin θ = y; cos θ = x; tan θ= y/x, or in terms of trigonometric identities: sin θ/cos θ. In the triangle, y may be replaced by sin θ and x may be replaced by cos θ; using pythagoras' theorem: 1= []²+[]² = [...] (sin θ)² + (cos θ)² = sin²θ + cos²θ.
An equilateral triangle with side lengths 2 is halved by perpendicular line that's the height, which creates triangles that are right angled. The length of the perpedicular bisector is √3; the triangles also have sides of 1 and 2, and angles of 30° and 60°. Thus sin 30° = [], cos 30° = [], tan 30° = [] sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3; sin 60° = [], cos 60° = [], tan 60° = [] sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
A right-angle isoceles triangle has two angles 45°, two sides 1 and the other side √2. It can give exact values of sine, cosine and tangent of 45°. sin 45° = 1/√2 (=√2/2), cos 45° = 1/√2 (=√2/2) and tan 45° = 1.
With trigonometric equations, where asked to solve f(nx) (e.g cos 2x = 0.81) in the range a°=<x=<b°: find all the solutions to nx, in the range na°=<nx<na°, then to find the sought solutions, have all the solutions to nx divided by n.
To solve sin²x = n: solve both sin x = +√n and sin x = -√n. That applies to other trigonometric functions.
With trigonometric functions, if f(x)=0= g(x) x h(x): for the full range of solutions to x: solve g(x) and h(x), putting both g(x) and h(x) equal to 0. Example: 2cos²x - cos x = 0 factorises to cos x(2cos x - 1)=0, so for solutions, solve both cos x = 0 and 2cos x - 1 = 0 (may be for a°=<x=<b°), though 2cos x - 1 = 0 can be simplified to cos x = 1/2. (Note: you may sometimes have quadratic equations with trigonometric functions [e.g 2cos²x + 3cos x - 2]).
With the graph of a^x, where a is positive and greater than or equal to 1, the [] the value of a, the steeper the graph higher--.
With the graph of a^-x, where a is positive and lower than or equal to 1, the [] the value of a, the steeper the graph lower--.
a^-x=(1/a)^x is a reflexion of a^x in the y-axis (as its f(-x)).
The graph a^x, moved one place to the right is a^x-1 (as it's f(x-1)).
The graph a^x, stretched along the x-axis by a scale factor of 1/n is a^nx (as it's f(nx)).
To simplify loga(m) + loga(n): let loga(m)=x, thus a^x= m, and loga(n)=y, thus a^y= n. a^x * a^y = [] = [] mn = a^(x+y), thus x+y=[]=[...] loga(mn)=loga(m)+loga(n).
To simplify loga(m) - loga(n): let loga(m)=x, thus a^x= m, and loga(n)=y, thus a^y= n. a^x / a^y = [] = [] m/n = a^(x-y), thus x-y=[]=[...] loga(m/n) = loga(m)-loga(n).
loga(m^n)=loga(m*m*...*m*m); using the addition rule: loga(m^n) = loga(m)+loga(m)+...+loga(m) = nloga(m).
Using the division rule: loga(1/m) = loga(1) - loga(m), which simplifies to [] - [] 0-loga(m)= -loga(m).
cos2x = [] = [] 2cos²x - 1 = cos²x - sin²x = 1 - 2sin²x.
When finding the area between the axes (or an axis) and a curve that crosses the axes through integration: add together the area of the sections, determined by when the curve crosses the x-axis. If the area of a section is negative, make it positive. Note: if finding the definite integral, such techniques are not needed; simply substitute in the limits and subtract - the definite integral can be negative or even zero!
When finding the area between the curve and the y-axis, integrate with respect to y, thus find (∫ ^b)a x dy (a ss), meaning that the subject of the equation to be integrated should be x. If the region is on both sides of the y-axis then treat (integrate) each part seperately, though in some cases doubling would suffice.
The trapezium rule is a numerical method for estimating integrals. It is most useful when, to an integral, there is no analytical answer and all that is needed is a number. It works by approximating the area under the curve by a series of trapezia, then adding the evaluated areas. The formula for the trapezium rule is (#a6#)where yp (p SS) = f(p), and the step size or the increment by which the values of x increase is h. For ease of calculation it may be a good idea to have the values of y tabulated.
In general, for the integral of concave functions, the trapezium method gives an underestimate, while it gives an overestimate for convex functions.
Created by: Toluo
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