Applications of Derivatives and Integration

Properties of definite integrals

A definite integral \int_a^b f(x)\,dx is a number fixed by the integrand and the two limits, so its value can be rewritten with algebraic rules before any antiderivative is found. Many NDA questions are solved this way, and these rules should be the first tools to try when direct integration looks messy.

Reversal and additivity. Interchanging the limits changes the sign: \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx. For any c between a and b, \int_a^b f=\int_a^c f+\int_c^b f. This splitting is the standard method whenever the integrand changes its formula, as with absolute values or the greatest integer function. For example, \int_0^3 |x-1|\,dx is split at x=1 to give \tfrac12+2=\tfrac52.

King property (the x\to a-x symmetry). For an integrable f, \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx, obtained by the substitution x\to a+b-x. On [0,a] this reads \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx. If the same integral I is written in both forms and the two forms are added, the sum 2I often simplifies to a constant. When the integrand is \frac{g(x)}{g(x)+g(a-x)}, the sum of the two forms is 1, so I=\frac{a}{2} whatever g is, provided the denominator never vanishes. If f(\pi-x)=-f(x) on [0,\pi], then I=-I, hence I=0. The integral \int_0^\pi \ln\tan\frac{x}{2}\,dx is of this type, because \tan\frac{\pi-x}{2}=\cot\frac{x}{2} and \ln\cot=-\ln\tan.

Even and odd functions on a symmetric interval. Splitting at 0 and substituting x\to -x in the left part gives \int_{-a}^{a}h(x)\,dx=\int_0^a[h(x)+h(-x)]\,dx. Consequently, if h is odd then \int_{-a}^{a}h=0, and if h is even then \int_{-a}^{a}h=2\int_0^a h. The same identity identifies the function g in a statement such as \int_0^a[f(x)+f(-x)]\,dx=\int_{-a}^a g(x)\,dx: here g=f. An odd-by-even product is odd, so a sine times a cosine squared integrates to zero over [-a,a]. For an even f and b>0, \int_{-a}^{a}\frac{f(x)}{1+b^x}\,dx=\int_0^a f(x)\,dx, because \frac{1}{1+b^x}+\frac{1}{1+b^{-x}}=1 pairs the two halves. For example, \int_{-2}^{2}(x^3+x^2)\,dx=\frac{16}{3}, since the odd part x^3 contributes nothing.

Periodicity. If T is a period of f, then \int_a^{a+T}f(x)\,dx is the same for every a, and over n periods \int_a^{a+nT}f=n\int_0^T f. The number of periods is the length of the interval divided by T. For example, \int_0^{6\pi}|\cos x|\,dx=6\cdot 2=12, since |\cos x| has period \pi and \int_0^{\pi}|\cos x|\,dx=2. The same counting applies to |\sin x| over [0,8\pi], which gives 16. Before counting, reduce the integrand to a form whose period is visible. The identity \sin^4x+\cos^4x=1-\frac12\sin^22x=\frac34+\frac14\cos4x shows that the integrand has period \frac{\pi}{2}; over [0,\pi] the cosine part integrates to zero, so the integral is \frac{3\pi}{4}. A function with period T has \varphi(a)=\int_a^{a+T}f constant in a. This is confirmed by the Leibniz rule, \varphi'(a)=f(a+T)-f(a)=0 for periodic f.

Fundamental theorem and inverse functions. For a function f with a continuous derivative, \int_a^b f'(x)\,dx=f(b)-f(a). The integral therefore depends only on the endpoint values, and a condition such as f(1)=f(4) forces the integral to vanish. Before integrating, simplify with e^{\ln u}=u for u>0, but check the sign of the base first. On (0,\frac{\pi}{2}) we have \cos x>0, so e^{\ln(\cos x)} simplifies to \cos x, which integrates to 1.

Sign functions. For x\neq 0, \frac{|x|}{x} equals -1 when x<0 and +1 when x>0. For a<b<0 this gives \int_a^b\frac{|x|}{x}\,dx=a-b. For a<0<b, splitting at 0 gives \int_a^0(-1)\,dx+\int_0^b 1\,dx=a+b.

Comparison of absolute values. Since |f(x)|\ge f(x) pointwise, \int_a^b|f|\ge\int_a^b f for a\le b, with equality exactly when f\ge 0 on the interval. The same reasoning gives \left|\int_a^b f\right|\le\int_a^b|f|. In particular, if f(x)=e^x is positive everywhere, then |f|=f and \int_a^b|f|=\int_a^b f for any limits, so the two integrals are equal.

Greatest integer function. The function [x] is constant on each interval [n,n+1) with n an integer, so an integrand built from [x] is handled by splitting the range at integers. On [-1,0) the value of [x] is -1, and on [1,2) the function reduces to a polynomial that can be integrated directly. The expression x-[x] is the fractional part of x, which has period 1. Hence \int_n^{n+1}(x-[x])\,dx=\int_0^1 x\,dx=\frac12 for every integer n.

Logarithmic integrals. Logarithm of a quotient splits as \ln\frac{A}{B}=\ln A-\ln B. The two standard values \int_0^1\ln x\,dx=-1 and \int_0^1\ln(1-x)\,dx=-1 (the second by u=1-x) then give \int_0^1\ln\left(\frac1x-1\right)dx=0. Results of the form c\ln k are combined by c\ln k=\ln k^{c}, and \ln\sqrt e=\frac12. For example, 3\ln2+\frac12=\ln(8\sqrt e), a form to recognise among answer options.

Integrals of x\,g(x). If g(\pi-x)=g(x) on [0,\pi], the substitution x\to\pi-x gives I=\int_0^\pi(\pi-x)g(x)\,dx. Adding the two forms yields 2I=\pi\int_0^\pi g, so I=\frac{\pi}{2}\int_0^\pi g(x)\,dx. Together with \int_0^\pi\frac{dx}{1+\cos^2x}=\frac{\pi}{\sqrt2}, this reduces such values to a power of \pi.

Named integrals. A value stated for a named integral such as I_2 is meaningful only together with its defining integral. The properties above can be applied only after that defining expression is identified, so any stated value should be verified against it before use.

Separable differential equations

A first-order differential equation is called separable when it can be written as \frac{dy}{dx}=g(x)\,h(y), so that dividing by h(y) places every y-term with dy and every x-term with dx. The method is to write \frac{dy}{h(y)}=g(x)\,dx, integrate both sides, and add one arbitrary constant c. The result is a general solution, which is a family of curves. An initial condition y(x_0)=y_0 fixes c and gives the particular solution.

The integrals that appear most often are \int\frac{dy}{y}=\ln|y| and \int\frac{dx}{x+1}=\ln|x+1|. Because a^{x}=e^{x\ln a}, the equation \frac{dy}{dx}=(\ln a)\,y separates to \frac{dy}{y}=(\ln a)\,dx and has solution y=Ca^{x}. An initial value y(0)=y_0 then gives C=y_0.

Exponential decay is the standard application. If the rate of decrease is proportional to the amount present, \frac{dN}{dt}=-kN gives N=N_0e^{-kt}. If half the substance remains after time T, then e^{-kT}=\frac{1}{2}, so k=\frac{\ln 2}{T}.

A velocity equation of the form \frac{dx}{dt}=g(x) separates to \int\frac{dx}{g(x)}=t. The limits must match the initial position and initial time, and the time needed to reach a position is the definite integral between those limits. Integrating \frac{1}{x+1} produces \ln(x+1), so a logarithm of a shifted variable often appears in the answer.

Some equations become separable only after simplification. In M\,dx+N\,dy=0, after collecting terms, the coefficients should factor into a function of x alone and a function of y alone. Identities such as 1+\cos x=2\cos^{2}\frac{x}{2} and 1-\cos x=2\sin^{2}\frac{x}{2} help in this step, and then \int\sec^{2}\frac{x}{2}\,dx=2\tan\frac{x}{2} is applied.

A relation F(x,y)=c is checked by implicit differentiation, F_x+F_y\frac{dy}{dx}=0, and the resulting slope must match the given equation. Any expression containing \ln\frac{dy}{dx} requires \frac{dy}{dx}>0, so a candidate with a non-positive slope cannot satisfy it.

Key terms:

• general solution — the family containing one arbitrary constant. 예) y=Cx^{2} represents all parabolas with vertex at the origin, which is a family of curves.
• particular solution — the member fixed by an initial condition. 예) y(0)=y_0 in y=Ce^{kx} gives C=y_0.
• factorising in p=\frac{dy}{dx} — a quadratic in p such as p^{2}-xp=0 factorises as p(p-x)=0, so p=0 or p=x, and each factor gives its own family. A candidate is accepted only if its slope satisfies the whole equation.

Two solutions of \frac{dy}{dx}=g(x) differ only by an additive constant, so their graphs are vertical translates of each other. If the difference of two solutions is a nonzero constant, the curves never meet.

When the right-hand side contains a power of y, integration gives a power of y with a negative exponent, for example \int y^{-3}\,dy=-\frac{1}{2y^{2}}. The solution is then an implicit relation, the constant is fixed by the initial value, and the final expression is usually rearranged to a clean form without fractions.

If the right-hand side depends only on a combination such as x+y, the substitution v=x+y gives \frac{dv}{dx}=1+\frac{dy}{dx}, and the equation becomes separable in v. An integral of the form \int\frac{dv}{v^{2}+k^{2}} produces an arctangent, and reversing the substitution gives a tangent relation.

Separable questions also test supporting evaluation skills. A determinant with an entire row of zeros equals zero, and at x=0 the entries \sin 0=0 and \tan 0=0 can produce such a row.

The greatest integer function [x] gives the largest integer not exceeding x, so [-2.3]=-3 while [2.3]=2. Cosine is even, so \cos(-\theta)=\cos\theta, and a negative integer multiplier can be handled through this property.

Large angles are reduced by the period 2\pi. For example, \sin\frac{11\pi}{4}=\sin\frac{3\pi}{4}=\frac{1}{\sqrt{2}} and \cos\frac{7\pi}{2}=0.

For a function of the form f(x)=\sum c_n e^{nx}, the derivative is f'(x)=\sum n\,c_n e^{nx}, so f'(0)=\sum n\,c_n. Each exponential contributes its exponent times its coefficient at x=0.

Local maxima and minima

A function f has a local maximum at c when f(c)\ge f(x) for all x near c, and a local minimum when f(c)\le f(x) for all x near c. Local extrema occur only at critical points, where f'(c)=0 or f'(c) does not exist. A critical point need not be an extremum, so each candidate must be tested.

Key terms:

• stationary point — a point where f'(c)=0. It may be a maximum, a minimum, or neither. 예) For f(x)=(x-2)^{2} type behaviour, the derivative vanishes without a sign change, so no extremum occurs.
• local extremum value — the value f(c) at the extremum. It is compared only with nearby values, so a local maximum value can be smaller than a local minimum value elsewhere on the graph.

The first derivative test checks the sign of f' on both sides of c. A change from positive to negative gives a maximum, a change from negative to positive gives a minimum, and no sign change gives no extremum. In a statement claiming a sign change direction, the direction decides whether the point is a maximum or a minimum, and the direction must be checked numerically on each side.

The second derivative test uses f''(c) at a critical point. If f''(c)>0 the point is a local minimum, and if f''(c)<0 it is a local maximum. If f''(c)=0 the test is inconclusive, and the first derivative test must be used. For a cubic whose derivative factorises as x(3x+m), the two critical points are separated by the sign of f'' at each.

For a logarithmic function such as f(x)=\frac{x}{\ln x}, the quotient rule gives f'(x)=\frac{\ln x-1}{(\ln x)^{2}}, and f' changes sign from negative to positive at x=e, so e is a local minimum. The second derivative at that point is \frac{1}{e}, which is positive and confirms the result.

A point where f' does not exist can still be an extremum. For a function such as 1-(x-1)^{2/3}, the expression (x-1)^{2/3}\ge 0 gives f(x)\le f(1) for all x, so x=1 is a maximum even though f'(1) is undefined.

For piecewise functions and functions with corners such as |x|, the corner and the endpoints of each piece must be checked separately. For f(x)=|x|+1 on a closed interval, the corner at x=0 is a local minimum even though the derivative does not exist there, and the maximum over the interval is attained at the endpoints. Both endpoints may give the same maximum value.

A sum of distances \sum|x-a_i| is minimised at the median of the points a_i. For three points a\le b\le c, the minimum equals c-a and is attained at x=b, because |b-a|+|b-c|=c-a. The triangle inequality gives the lower bound |c-a| for the sum.

When y is defined by a relation such as y^{2}=k-m x^{2} with m>0, the squared term is nonnegative, so y^{2}\le k. The greatest y is the square root of k, reached when x=0.

For a cubic f, the derivative is a quadratic whose discriminant decides whether critical points exist. For f'(x)=3x^{2}+2x+k the discriminant is 4-12k. If f' has no real zeros and keeps one sign, f is strictly monotonic and has no local extremum, which requires the discriminant to be negative. The boundary case with zero discriminant gives a stationary point without a sign change, so the strict inequality is the usual key.

For a function of a parameter, the derivative at a fixed point is often a linear expression in that parameter. If y=2f(x)+ax-b has f'(0)=0, then y'(0)=a, so a stationary point at x=0 requires a=0, while an additive constant only shifts the graph vertically and does not affect the stationary point.

For an expression such as p^{2}\sec^{2}x+q^{2}\csc^{2}x, substituting T=\tan^{2}x turns it into p^{2}(1+T)+q^{2}\left(1+\frac{1}{T}\right). Setting the derivative with respect to T to zero gives T=\frac{q}{p}, which is the minimising value.

To count the extreme values of a function such as \sin 4x+2x on an open interval, solve f'(x)=0 inside the interval and check the sign change at each root. The equation \cos\theta=-\frac{1}{2} has solutions \theta=\frac{2\pi}{3} and \frac{4\pi}{3} in (0,2\pi), and these give the points where the extrema of such a function can occur.

When the domain is restricted, only critical points inside the domain count. For a derivative 16x^{2}-8bx+1 with x>0, the discriminant 64(b^{2}-1) decides whether real roots exist, and the sum and product of the roots decide whether those roots are positive. A positive product means both roots have the same sign, and a positive sum then means both are positive.

Area under a curve by definite integral

The definite integral \int_a^b f(x)\,dx gives the area between a continuous curve y=f(x), the x-axis and the lines x=a, x=b, provided f(x)\ge 0 on [a,b]. Where the curve lies below the x-axis, the signed integral is negative, so an area question cannot simply evaluate the integral. Split the interval at each zero of f and integrate |f| over every piece. The total area is \int_a^b |f(x)|\,dx. For instance, f(x)=x^3 on [-1,2] has signed integral \tfrac{15}{4}, but its area is \tfrac{1}{4}+4=\tfrac{17}{4}, because the part below the axis is counted as positive.

When a region is bounded by two curves, the area is \int_a^b (f_{\text{upper}}-f_{\text{lower}})\,dx. The limits a and b are the x-coordinates of the intersection points, and the upper curve is identified by testing a point between them. For y=2x and y=x^2, the intersections are at x=0 and x=2, so the area is \int_0^2 (2x-x^2)\,dx=\tfrac{4}{3}.

For lines y=kx and y=mx with k>m, bounded by the vertical line x=a, the enclosed area is \tfrac{(k-m)a^2}{2}. When a parameter in such lines is unknown, equate this area to the given value and solve. The sign of the parameter decides which line lies above, so check the sign before choosing the integrand.

When the boundary is given as x=g(y), integrate with respect to y. A region bounded by x=|y| and a vertical line x=c is a triangle with base 2c and height c, so its area is c^2.

The graph y=\sqrt{r^2-x^2} with y\ge 0 is the upper semicircle of x^2+y^2=r^2, and its area is \tfrac{1}{2}\pi r^2. A circular segment whose arc subtends a central angle \theta has area \tfrac{r^2}{2}(\theta-\sin\theta), which is the sector \tfrac{r^2\theta}{2} minus the triangle \tfrac{r^2\sin\theta}{2}. The two segments cut off by one chord together fill the disc, so A_1+A_2=\pi r^2, and the major segment equals \pi r^2 minus the minor segment.

A region bounded by an arc, a line and the axis is split at the intersection point into a triangular part and an arc part. Each part is found separately and the results are added. A region can also be found by subtraction: the area of a known shape minus the area under a curve over the same interval. For example, a quarter-disc of radius a has area \tfrac{\pi a^2}{4}, and removing \int_0^\pi \sin x\,dx=2 from it leaves the required region.

For y=[x], the greatest integer function is constant n on each [n,n+1). Its graph is a set of horizontal steps, so the area is a sum of rectangles, each with width times absolute height. For example, \int_{0.5}^{2.5}[x]\,dx with [x]=0,1,2 on successive pieces gives 0+1+1=2.

• bounded by — enclosed by all the listed curves and lines, forming a closed region with finite area. e.g. The region bounded by y=x, y=0 and x=1 is a triangle of area \tfrac{1}{2}.
• in the first quadrant — restrict to x\ge 0, y\ge 0, so the limits are clipped to that quadrant. e.g. The area in the first quadrant under y=\sqrt{x} up to x=4 is \tfrac{16}{3}.
• between two curves — upper curve minus lower curve, with limits at the intersection points.

Order and degree of differential equations

The order of a differential equation is the order of the highest derivative occurring in it. Equivalently, it equals the number of essential arbitrary constants in the general solution, because each differentiation of the family removes one independent constant. A family such as y=Ae^x+Be^{-x} has two constants and gives a second-order equation. A constant that appears in several places but remains a single parameter counts only once, so a family with one constant c gives a first-order equation. A family of ellipses with axes along the coordinate axes, \tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1, has two arbitrary constants and therefore order 2.