Module 1 · NDA Mathematics (Paper 1)

Calculus

Limits, continuity, differentiation, integration, differential equations.
Limits · Differentiation · Integration · Differential Equations
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1. Limits

The concept of limit is the foundation of calculus. For a function f(x), we say limx→a f(x) = L if f(x) gets arbitrarily close to L when x is sufficiently close to a (but not equal to a). Limits are used to define derivatives, continuity, and integrals.

Basic Limits & Indeterminate Forms

When evaluating limits, we may encounter forms that are not immediately determinate. These require algebraic manipulation, standard limit formulas, or L'Hopital's Rule.

Indeterminate Forms

The following seven forms are indeterminate. Further analysis is required to evaluate limits that produce these forms. When you directly substitute the limiting value and obtain one of these forms, it means the limit cannot be determined without additional work.

Rationalization Method

When the limit involves square roots, rationalizing the numerator or denominator often resolves the 0/0 form. Multiply by the conjugate to eliminate the radical.

Example 1 | NDA Practice
Evaluate: limx→0 (√(1 + x) − 1)/x
Solution: Multiply numerator and denominator by (√(1 + x) + 1): = limx→0 [(1+x) − 1]/[x(√(1+x) + 1)] = limx→0 x/[x(√(1+x) + 1)] = limx→0 1/(√(1+x) + 1) = 1/(1+1) = 1/2.
⚡ NDA Shortcut: Conjugate Multiplication
When you see √(a + x) − √a or similar forms, always multiply by the conjugate. The formula (√A − √B)(√A + √B) = A − B eliminates the radicals. This is a very common technique in NDA limit problems.
Indeterminate FormExampleTechnique
0/0limx→0 (sin x)/xFactorisation, standard limits, L'Hopital
∞/∞limx→∞ ex/xDivide by highest power, L'Hopital
0 × ∞limx→0+ x ln xRewrite as 0/0 or ∞/∞
∞ − ∞limx→0 (1/x − 1/sin x)Combine fractions, rationalise
00limx→0+ xxTake ln, use L'Hopital
0limx→∞ x1/xTake ln, use L'Hopital
1limx→0 (1 + x)1/xUse standard exponential limit
Example 2 | NDA 2019
Evaluate: limx→0 (1 − cos 2x)/x2
Solution: Using 1 − cos 2x = 2 sin2x, we get limx→0 2 sin2x / x2 = 2 × (limx→0 sin x / x)2 = 2 × 12 = 2.
Example 3 | NDA 2020
Evaluate: limx→2 (x2 − 4)/(x − 2)
Solution: Factorise numerator: x2 − 4 = (x − 2)(x + 2). Cancel (x − 2) to get limx→2 (x + 2) = 4.
⚡ NDA Shortcut: 0/0 Forms
For rational 0/0 limits, always try factorisation first. If the numerator and denominator are polynomials, check if (x − a) is a common factor before applying L'Hopital. This saves time in NDA exams.

Standard Limits

Memorising standard limits is essential for quick evaluation in NDA exams. The following tables list the most frequently tested limits.

Trigonometric Limits

Limit FormulaResult
limx→0 sin x / x1
limx→0 tan x / x1
limx→0 (1 − cos x) / x21/2
limx→0 sin−1 x / x1
limx→0 tan−1 x / x1
limx→a (sin x − sin a) / (x − a)cos a
Example 4 | NDA 2018
Evaluate: limx→0 (tan 3x)/x
Solution: Write as 3 × (tan 3x)/(3x). Using limθ→0 tan θ / θ = 1, we get 3 × 1 = 3.

Exponential & Logarithmic Limits

Limit FormulaResult
limx→0 (ex − 1) / x1
limx→0 (ax − 1) / xln a
limx→0 (1 + x)1/xe
limx→∞ (1 + 1/x)xe
limx→0 ln(1 + x) / x1
limx→0 (eax − 1) / xa
Example 5 | NDA 2019
Evaluate: limx→0 (e2x − 1)/x
Solution: Using the standard limit limx→0 (eax − 1)/x = a, we get 2. Alternatively, apply L'Hopital's rule: (2e2x)/1 → 2.
⚡ NDA Shortcut: Exponential Limits
For limits of the form limx→0 (eax − 1)/(bx), the answer is a/b. Similarly, limx→0 (ax − 1)/x = ln a. These direct formulas save 2−3 steps.

Series Expansions for Limits

Standard series expansions are powerful tools for evaluating limits, especially when L'Hopital's rule becomes tedious. These expansions represent functions as infinite polynomials around x = 0.

FunctionSeries Expansion (Maclaurin)
ex1 + x + x2/2! + x3/3! + ...
sin xx − x3/3! + x5/5! − ...
cos x1 − x2/2! + x4/4! − ...
tan xx + x3/3 + 2x5/15 + ...
ln(1 + x)x − x2/2 + x3/3 − x4/4 + ...
(1 + x)n1 + nx + n(n−1)x2/2! + ...
sin−1 xx + x3/6 + 3x5/40 + ...
tan−1 xx − x3/3 + x5/5 − ...
Example 6 | NDA Practice
Evaluate using series: limx→0 (sin x − x)/x3
Solution: Using sin x = x − x3/3! + x5/5! − ..., we have sin x − x = −x3/6 + x5/120 − ... Dividing by x3: (sin x − x)/x3 = −1/6 + x2/120 − ... As x→0, the limit is −1/6.
⚡ NDA Shortcut: Series vs L'Hopital
For limits involving combinations of sin, cos, ex, and ln(1+x), series expansion often gives the answer in one step compared to repeated L'Hopital. Memorise the first 2−3 terms of ex, sin x, cos x, and ln(1+x).

L'Hopital's Rule

If limx→a f(x)/g(x) gives 0/0 or ∞/∞, then limx→a f(x)/g(x) = limx→a f'(x)/g'(x), provided the limit exists. The rule can be applied repeatedly until a determinate form is obtained.

ConditionActionExample
0/0 formDifferentiate numerator and denominator separatelylimx→0 (ex − 1)/x = limx→0 ex/1 = 1
∞/∞ formDifferentiate numerator and denominator separatelylimx→∞ x/ex = limx→∞ 1/ex = 0
0 × ∞ formRewrite as 0/0 or ∞/∞ firstlimx→0+ x ln x = limx→0+ (ln x)/(1/x) = 0
Example 7 | NDA 2020
Evaluate: limx→0 (ex − 1 − x)/x2
Solution: This is 0/0 form. Apply L'Hopital's rule: limx→0 (ex − 1)/(2x). Still 0/0. Apply again: limx→0 ex/2 = 1/2. So the answer is 1/2.
⚡ NDA Shortcut: L'Hopital Verification
Before applying L'Hopital, always verify that the form is truly 0/0 or ∞/∞. If not, L'Hopital gives wrong answers. In NDA exams, check by substituting the limiting value first.

2. Continuity & Differentiability

Continuity at a Point

A function f(x) is continuous at x = a if limx→a f(x) = f(a). This requires three conditions:

  1. f(a) is defined (a belongs to the domain)
  2. limx→a f(x) exists (LHL = RHL)
  3. limx→a f(x) = f(a)

If any condition fails, f is discontinuous at x = a.

Properties of Continuous Functions

Continuous functions have several important properties that are frequently tested in NDA exams:

  • The sum, difference, product, and quotient (with non-zero denominator) of continuous functions are continuous.
  • Polynomials, trigonometric functions, exponential functions, and logarithmic functions are continuous on their domains.
  • The composition of two continuous functions is continuous.
  • If f is continuous on [a, b] and f(a) · f(b) < 0, then there exists c ∈ (a, b) such that f(c) = 0 (Intermediate Value Theorem).
  • A continuous function on a closed interval attains its maximum and minimum values (Extreme Value Theorem).
Example 8 | NDA 2019
Check continuity of f(x) = |x − 3| at x = 3.
Solution: LHL = limx→3 |x − 3| = 0. RHL = limx→3+ |x − 3| = 0. f(3) = 0. Since LHL = RHL = f(3), f is continuous at x = 3.

Types of Discontinuities

TypeDescriptionExample
Removablelimx→a f(x) exists but ≠ f(a) or f(a) undefinedf(x) = (x2 − 1)/(x − 1) at x = 1
JumpLHL and RHL exist but are not equalf(x) = { x, x < 0; x + 1, x ≥ 0 } at x = 0
InfiniteAt least one one-sided limit is ∞ or −∞f(x) = 1/x at x = 0
OscillatoryLimit does not exist due to oscillationf(x) = sin(1/x) at x = 0

Differentiability

A function f is differentiable at x = a if the derivative f'(a) = limh→0 [f(a + h) − f(a)]/h exists (LHD = RHD). Differentiability implies continuity, but the converse is not true.

Example 9 | NDA 2020
Is f(x) = |x − 3| differentiable at x = 3?
Solution: LHD = limh→0 (|3+h−3| − 0)/h = limh→0 |h|/h = −1. RHD = limh→0+ |h|/h = 1. Since LHD ≠ RHD, f is not differentiable at x = 3, though it is continuous.
⚡ NDA Shortcut: Continuity vs Differentiability
For NDA, remember: all differentiable functions are continuous, but not all continuous functions are differentiable. Modulus functions (|x|) and functions with sharp corners are classic examples of continuous but non-differentiable functions.

3. Differentiation

Basic Rules & Formulas

The derivative represents the rate of change. Geometrically, f'(x) is the slope of the tangent at point (x, f(x)).

FunctionDerivative
xnnxn−1
exex
axax ln a
ln x1/x
loga x1/(x ln a)
sin xcos x
cos x−sin x
tan xsec2 x
sec xsec x tan x
cosec x−cosec x cot x
cot x−cosec2 x
sin−1 x1/√(1 − x2)
cos−1 x−1/√(1 − x2)
tan−1 x1/(1 + x2)

Chain Rule, Product Rule, Quotient Rule

Chain Rule: If y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx).

Product Rule: (uv)' = u'v + uv'.

Quotient Rule: (u/v)' = (u'v − uv')/v2.

Example 10 | NDA 2018
Differentiate y = sin(ln x) with respect to x.
Solution: Using chain rule, dy/dx = cos(ln x) × d/dx(ln x) = cos(ln x)/x.
⚡ NDA Shortcut: Chain Rule
For nested functions, differentiate from outside to inside. In NDA exams, chain rule problems often involve trigonometric, exponential, and logarithmic compositions. Remember: derivative of f(g(x)) = f'(g(x)) × g'(x).

Implicit Differentiation

When y cannot be expressed explicitly as a function of x, differentiate both sides of the equation with respect to x, treating y as a function of x and using the chain rule (dy/dx appears).

Example 11 | NDA 2019
Find dy/dx if x3 + y3 = 3xy.
Solution: Differentiate both sides: 3x2 + 3y2(dy/dx) = 3y + 3x(dy/dx). Rearranging: 3y2(dy/dx) − 3x(dy/dx) = 3y − 3x2. So dy/dx = (y − x2)/(y2 − x).

Logarithmic Differentiation

Used when the function is of the form [f(x)]g(x) or a product/quotient of many functions. Take natural log on both sides and then differentiate.

Example 12 | NDA 2020
Find dy/dx if y = xx.
Solution: Take ln on both sides: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + x(1/x) = ln x + 1. So dy/dx = y(ln x + 1) = xx(ln x + 1).
⚡ NDA Shortcut: Logarithmic Differentiation
For y = [f(x)]g(x), the formula is: dy/dx = [f(x)]g(x) × [g'(x) ln f(x) + g(x) f'(x)/f(x)]. Memorise this to avoid re-deriving each time.

Second Order Derivatives

The second derivative f''(x) or d2y/dx2 is the derivative of the first derivative. It represents the rate of change of slope and is used to determine concavity and classify critical points.

If f''(x) > 0 on an interval, the function is concave up (holds water). If f''(x) < 0, the function is concave down (spills water). Points where concavity changes are called points of inflection (f''(x) = 0).

Successive Differentiation (nth Derivatives)

Successive differentiation involves finding higher-order derivatives. Some standard formulas for nth derivatives are useful in NDA exams:

f(x)f(n)(x)
xmm(m−1)...(m−n+1) xm−n
eaxan eax
sin(ax + b)an sin(ax + b + nπ/2)
cos(ax + b)an cos(ax + b + nπ/2)
ln(ax + b)(−1)n−1 (n−1)! an/(ax + b)n
Example 13 | NDA 2018
If y = ex sin x, find d2y/dx2.
Solution: dy/dx = ex sin x + ex cos x = ex(sin x + cos x). d2y/dx2 = ex(sin x + cos x) + ex(cos x − sin x) = 2ex cos x.

4. Applications of Derivatives

Rate of Change

The derivative dy/dx gives the rate of change of y with respect to x. In NDA problems, this is often used with time as the independent variable: dy/dt = (dy/dx)(dx/dt). Related rates problems involve finding the rate at which one quantity changes with respect to time given the rate of another related quantity.

Example 14 | NDA 2019
The radius of a sphere is increasing at 2 cm/s. Find the rate of increase of volume when radius = 3 cm.
Solution: V = (4/3)πr3. dV/dt = 4πr2 × dr/dt = 4π(9) × 2 = 72π cm3/s.

Increasing/Decreasing Functions

A function f(x) is increasing if f'(x) > 0 and decreasing if f'(x) < 0. If f'(x) = 0, the function may have a stationary point.

Example 15 | NDA 2020
Find the interval where f(x) = x3 − 3x + 2 is increasing.
Solution: f'(x) = 3x2 − 3 = 3(x − 1)(x + 1). f'(x) > 0 when x < −1 or x > 1. So f is increasing on (−∞, −1) ∪ (1, ∞).
⚡ NDA Shortcut: Monotonicity
To quickly determine increasing/decreasing intervals, factorise f'(x) and check the sign using the number line test. For quadratic f'(x), the sign alternates between roots.

Tangents & Normals

Tangent: Equation at (x0, y0) is y − y0 = f'(x0)(x − x0). The slope of the tangent gives the instantaneous rate of change at that point.

Normal: Equation is y − y0 = −1/f'(x0) × (x − x0), provided f'(x0) ≠ 0. The normal is perpendicular to the tangent (product of slopes = −1).

Lengths: The length of tangent = |y0 √(1 + (f'(x0))2)/f'(x0)|. The length of normal = |y0 √(1 + (f'(x0))2)|. The sub-tangent = |y0/f'(x0)|. The sub-normal = |y0 f'(x0)|.

Example 16 | NDA 2018
Find the equation of tangent to y = x2 at (1, 1).
Solution: dy/dx = 2x, so slope at (1,1) = 2. Equation: y − 1 = 2(x − 1) ⇒ y = 2x − 1.

Mean Value Theorem & Rolle's Theorem

Rolle's Theorem: If f is continuous on [a, b] and differentiable on (a, b), and f(a) = f(b), then there exists c ∈ (a, b) such that f'(c) = 0.

Lagrange's Mean Value Theorem (LMVT): If f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that f'(c) = [f(b) − f(a)]/(b − a).

Example 17 | NDA 2019
Verify Rolle's Theorem for f(x) = x2 − 4x + 3 on [1, 3].
Solution: f(1) = 1 − 4 + 3 = 0, f(3) = 9 − 12 + 3 = 0. So f(1) = f(3). f'(x) = 2x − 4. Setting f'(c) = 0 gives 2c − 4 = 0 ⇒ c = 2 ∈ (1, 3). Hence Rolle's Theorem is verified. c = 2.
⚡ NDA Shortcut: MVT Applications
In NDA exams, Rolle's Theorem is tested by verifying f(a)=f(b) and finding c where f'(c)=0. LMVT is tested by finding c such that f'(c) equals the average rate of change. Both are used to prove inequalities.

Maxima & Minima

First Derivative Test: Find critical points (f'(x) = 0). If f' changes from + to −, it is a local maximum; if − to +, local minimum.

Second Derivative Test: At a critical point, if f''(x) > 0, local minima; if f''(x) < 0, local maxima; if f''(x) = 0, test fails.

ConditionConclusion
f'(c) = 0 and f''(c) > 0Local minimum at x = c
f'(c) = 0 and f''(c) < 0Local maximum at x = c
f'(c) = 0 and f''(c) = 0Test inconclusive (use first derivative test)
f'(x) > 0 on both sides of cPoint of inflection
Example 18 | NDA 2020
Find the local maxima/minima of f(x) = 2x3 − 9x2 + 12x − 3.
Solution: f'(x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2) = 6(x − 1)(x − 2). Critical points: x = 1, 2. f''(x) = 12x − 18. f''(1) = −6 < 0 ⇒ local maximum at x = 1, f(1) = 2. f''(2) = 6 > 0 ⇒ local minimum at x = 2, f(2) = 1.
⚡ NDA Shortcut: Maxima/Minima
For NDA objective questions, after finding critical points, instantly check f''(x) sign. If only one critical point exists and it is a maximum, it is the global maximum for the given interval. Similarly for minimum.

Approximations & Differentials

Using differentials: f(x + Δx) ≈ f(x) + f'(x)Δx. This is useful for approximate calculations when Δx is small. The error in approximation is proportional to (Δx)2.

Absolute Error: |Δy| ≈ |f'(x) Δx|. Relative Error: |Δy/y| ≈ |f'(x)/f(x) Δx|. Percentage Error: (|Δy/y|) × 100%.

Example 19 | NDA 2019
Find the approximate value of √25.2.
Solution: Let f(x) = √x, f'(x) = 1/(2√x). Take x = 25, Δx = 0.2. f(25.2) ≈ f(25) + f'(25) × 0.2 = 5 + 1/(10) × 0.2 = 5 + 0.02 = 5.02.

5. Integration

Indefinite Integrals

Integration is the inverse process of differentiation. The indefinite integral ∫ f(x) dx = F(x) + C, where F'(x) = f(x) and C is the constant of integration. Every differentiation formula gives rise to a corresponding integration formula.

MethodWhen to UseExample
Direct FormulaStandard forms (power, trig, exponential)∫ x2 dx = x3/3 + C
SubstitutionComposite functions, integrand contains f'(x)∫ 2x cos(x2) dx
Integration by PartsProduct of two dissimilar functions∫ x ex dx
Partial FractionsRational functions P(x)/Q(x)∫ 1/(x2−4) dx
Trigonometric IdentitiesPowers of sin, cos, tan∫ sin2x dx = ∫ (1−cos 2x)/2 dx

Basic Formulas

IntegralResult
∫ xn dxxn+1/(n+1) + C, n ≠ −1
∫ (1/x) dxln |x| + C
∫ ex dxex + C
∫ ax dxax/ln a + C
∫ sin x dx−cos x + C
∫ cos x dxsin x + C
∫ sec2 x dxtan x + C
∫ cosec2 x dx−cot x + C
∫ sec x tan x dxsec x + C
∫ cosec x cot x dx−cosec x + C
∫ 1/(1 + x2) dxtan−1 x + C
∫ 1/√(1 − x2) dxsin−1 x + C

Substitution Method

If the integrand contains a composite function, substitute u = g(x) so that du = g'(x) dx and the integral simplifies to ∫ f(u) du. Common substitutions include u = sin x, cos x, tan x, ex, or algebraic expressions under radicals.

Integrand ContainsSubstitutionResult
√(a2 − x2)x = a sin θa cos θ dθ
√(x2 + a2)x = a tan θa sec2θ dθ
√(x2 − a2)x = a sec θa sec θ tan θ dθ
a2 − x2 in denominatorx = a sin θa cos θ dθ
a2 + x2 in denominatorx = a tan θa sec2θ dθ
√(a − x)/(a + x)x = a cos 2θTrigonometric simplification
Example 20 | NDA 2020
Evaluate: ∫ 2x cos(x2) dx
Solution: Let u = x2, du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x2) + C.
⚡ NDA Shortcut: Substitution Recognition
Look for integrands where one part is the derivative of another. For example, if the integrand contains f'(x) × g(f(x)), substitute u = f(x). Common patterns: x cos(x2), ex sin(ex), etc.

Integration by Parts

Using the formula: ∫ u dv = uv − ∫ v du. Choose u and dv using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) where u is chosen from the first category.

Example 21 | NDA 2019
Evaluate: ∫ x ex dx
Solution: Using integration by parts with u = x, dv = exdx: du = dx, v = ex. ∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C.
⚡ NDA Shortcut: LIATE Rule
LIATE gives the priority for u in integration by parts: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential. The remaining factor becomes dv. For products like xn ex, choose u = xn.

Partial Fractions

Used to integrate rational functions P(x)/Q(x). Decompose the fraction into simpler terms based on the factors of Q(x).

Factor in DenominatorPartial Fraction Form
(ax + b)A/(ax + b)
(ax + b)nA1/(ax+b) + A2/(ax+b)2 + ... + An/(ax+b)n
ax2 + bx + c (non-repeated)(Ax + B)/(ax2 + bx + c)
Example 22 | NDA 2020
Evaluate: ∫ 1/(x2 − 4) dx
Solution: Using partial fractions: 1/(x2 − 4) = 1/[(x−2)(x+2)] = (1/4)[1/(x−2) − 1/(x+2)]. Integrating: (1/4)[ln|x−2| − ln|x+2|] + C = (1/4) ln|(x−2)/(x+2)| + C.

Integration Using Trigonometric Identities

Trigonometric identities simplify integrals involving products of sine and cosine, powers of trigonometric functions, and rational trigonometric expressions.

  • sin2x = (1 − cos 2x)/2, cos2x = (1 + cos 2x)/2
  • sin 3x = 3 sin x − 4 sin3x, cos 3x = 4 cos3x − 3 cos x
  • 2 sin A cos B = sin(A+B) + sin(A−B)
  • 2 cos A cos B = cos(A+B) + cos(A−B)
  • 2 sin A sin B = cos(A−B) − cos(A+B)
Example 23 | NDA Practice
Evaluate: ∫ sin 3x cos 2x dx
Solution: Using 2 sin A cos B = sin(A+B) + sin(A−B): ∫ sin 3x cos 2x dx = (1/2) ∫ (sin 5x + sin x) dx = (1/2)[−cos 5x/5 − cos x] + C = −cos 5x/10 − cos x/2 + C.

Integration of Special Functions

IntegralResult
∫ 1/(x2 + a2) dx(1/a) tan−1(x/a) + C
∫ 1/(x2 − a2) dx(1/2a) ln |(x−a)/(x+a)| + C
∫ 1/√(a2 − x2) dxsin−1(x/a) + C
∫ 1/√(x2 + a2) dxsinh−1(x/a) + C
∫ 1/√(x2 − a2) dxln |x + √(x2 − a2)| + C
∫ √(a2 − x2) dx(x/2)√(a2−x2) + (a2/2) sin−1(x/a) + C
∫ tan x dxln |sec x| + C
∫ cot x dxln |sin x| + C
∫ sec x dxln |sec x + tan x| + C
∫ cosec x dxln |cosec x − cot x| + C
∫ eax sin(bx) dxeax(a sin bx − b cos bx)/(a2 + b2) + C
⚡ NDA Shortcut: Integration Speed Tricks
For ∫ eax sin(bx) dx and ∫ eax cos(bx) dx, use the direct formula: eax(a sin bx − b cos bx)/(a2+b2) and eax(a cos bx + b sin bx)/(a2+b2) respectively. No need to integrate by parts twice.

Reduction Formulas (for NDA Advanced Problems)

Reduction formulas express an integral involving powers of functions in terms of lower powers. These are useful for integrals like ∫ sinnx dx and ∫ cosnx dx.

Reduction FormulaResult
∫ sinnx dx−(1/n) sinn−1x cos x + ((n−1)/n) ∫ sinn−2x dx
∫ cosnx dx(1/n) cosn−1x sin x + ((n−1)/n) ∫ cosn−2x dx
∫ tannx dxtann−1x/(n−1) − ∫ tann−2x dx
∫ secnx dx(1/(n−1)) secn−2x tan x + ((n−2)/(n−1)) ∫ secn−2x dx
Example 24 | NDA Practice
Use the reduction formula to find ∫ sin2x dx.
Solution: Using the reduction formula with n=2: ∫ sin2x dx = −(1/2) sin x cos x + (1/2) ∫ sin0x dx = −(1/2) sin x cos x + x/2 + C = (x − sin x cos x)/2 + C.
⚡ NDA Shortcut: Even Powers of sin/cos
For sin2x and cos2x, use the double-angle identities directly: sin2x = (1−cos 2x)/2 and cos2x = (1+cos 2x)/2. This is faster than the reduction formula for n=2.

Definite Integrals

The definite integral ∫ab f(x) dx = F(b) − F(a), where F'(x) = f(x). It represents the signed area under the curve y = f(x) from x = a to x = b.

Properties of Definite Integrals

PropertyFormula
Reversal of limitsab f(x) dx = −∫ba f(x) dx
Zero widthaa f(x) dx = 0
Additivityab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx
King's propertyab f(x) dx = ∫ab f(a + b − x) dx
Odd function−aa f(x) dx = 0 if f is odd
Even function−aa f(x) dx = 2 ∫0a f(x) dx if f is even
Periodic function0nT f(x) dx = n ∫0T f(x) dx if period = T
I = ∫ab f(x) dx = ∫ab f(a+b−x) dxAdding: 2I = ∫ab [f(x)+f(a+b−x)] dx
Example 25 | NDA 2020
Evaluate: ∫0π/2 sin x/(sin x + cos x) dx
Solution: Let I = ∫0π/2 sin x/(sin x + cos x) dx. Using King's property: I = ∫0π/2 cos x/(cos x + sin x) dx. Adding: 2I = ∫0π/2 1 dx = π/2. So I = π/4.
⚡ NDA Shortcut: King's Property
For definite integrals with limits a and b, replace x by a+b−x and add the two integrals. This is a powerful trick for NDA problems involving sin and cos ratios.

Inequalities & Estimation of Definite Integrals

Several important inequalities help estimate definite integrals without evaluating them exactly:

  • If m ≤ f(x) ≤ M on [a, b], then m(b − a) ≤ ∫ab f(x) dx ≤ M(b − a).
  • Cauchy-Schwarz Inequality: (∫ab f(x)g(x) dx)2 ≤ ∫ab [f(x)]2 dx · ∫ab [g(x)]2 dx.
  • Triangle Inequality: |∫ab f(x) dx| ≤ ∫ab |f(x)| dx.
  • Mean Value Theorem for Integrals: There exists c ∈ [a, b] such that ∫ab f(x) dx = f(c)(b − a).
Example 26 | NDA Practice
Find the average value of f(x) = x2 on [0, 2].
Solution: Average value = (1/(b−a)) ∫ab f(x) dx = (1/2) ∫02 x2 dx = (1/2)[x3/3]02 = (1/2)(8/3) = 4/3.
⚡ NDA Shortcut: Average Value
Average value of a function on [a, b] is (1/(b−a)) times the definite integral. For NDA problems asking "average value", use this formula directly. Common functions have simple averages.

Area Under Curves

The area bounded by y = f(x), x-axis, x = a, x = b is given by ∫ab |f(x)| dx. The area between two curves y = f(x) and y = g(x) is ∫ab |f(x) − g(x)| dx.

Example 27 | NDA 2018
Find the area bounded by the curves y = x2 and y = 2x.
Solution: Intersection: x2 = 2x ⇒ x2 − 2x = 0 ⇒ x(x − 2) = 0, so x = 0, 2. Area = ∫02 (2x − x2) dx = [x2 − x3/3]02 = (4 − 8/3) − (0) = 4/3 sq. units.
⚡ NDA Shortcut: Area Between Curves
Quickly identify the upper and lower curves by testing a point between intersection points. The upper curve minus lower curve gives the positive integrand. Common NDA areas: parabola − line, circle − parabola.

6. Differential Equations

Order & Degree

Order is the highest derivative present. Degree is the power of the highest derivative after removing radicals and fractions. A differential equation is linear if the dependent variable and its derivatives appear in the first degree. If the DE involves products or powers of the dependent variable or its derivatives, it is non-linear.

Example 28 | NDA 2019
Find the order and degree of (d2y/dx2)3 + (dy/dx)2 + y = 0.
Solution: Highest derivative is d2y/dx2 (order 2). Its power is 3 (degree 3). There are no radicals or fractions. So order = 2, degree = 3.

Variable Separable

A DE of the form dy/dx = f(x)g(y) can be solved by separating variables: ∫ dy/g(y) = ∫ f(x) dx.

Example 29 | NDA 2020
Solve: dy/dx = y/x
Solution: Separate variables: dy/y = dx/x. Integrate: ln|y| = ln|x| + C ⇒ ln|y| = ln|x| + ln K ⇒ y = Kx. So y = Cx, where C is a constant.
⚡ NDA Shortcut: Variable Separable
In NDA exams, variable separable DEs often come in the form dy/dx = f(x)/g(y). Rearrange to g(y) dy = f(x) dx and integrate directly. Always add the constant of integration on one side.

Homogeneous DE

A DE of the form dy/dx = F(y/x) is homogeneous. Substitute y = vx, then dy/dx = v + x(dv/dx) to convert it to a variable separable equation.

Example 30 | NDA 2018
Solve: (x2 + y2) dx = 2xy dy
Solution: Rewrite as dy/dx = (x2 + y2)/(2xy). Divide numerator and denominator by x2: dy/dx = (1 + (y/x)2)/(2(y/x)). Substitute y = vx: v + x(dv/dx) = (1 + v2)/(2v). Simplify: x(dv/dx) = (1 + v2)/(2v) − v = (1 − v2)/(2v). Separate: (2v/(1−v2)) dv = dx/x. Integrate: −ln|1−v2| = ln|x| + C ⇒ 1/(1−v2) = Kx ⇒ x2 − y2 = Cx.

Linear DE

A first-order linear DE is of the form dy/dx + P(x)y = Q(x). The integrating factor (IF) is e∫ P dx. The solution is y(IF) = ∫ Q(IF) dx + C. Linear DEs appear frequently in NDA exams and can be identified by the dependent variable y and its derivative dy/dx appearing only in the first degree without any product terms.

Example 31 | NDA 2020
Solve: dy/dx + (2/x)y = x
Solution: Here P = 2/x, Q = x. IF = e∫ (2/x) dx = e2 ln x = x2. Solution: y × x2 = ∫ x × x2 dx = ∫ x3 dx = x4/4 + C. So y = x2/4 + C/x2.
⚡ NDA Shortcut: Linear DE
The integrating factor IF = e∫ P dx. Once found, the solution is always y(IF) = ∫ Q(IF) dx + C. For NDA, common P forms are 2/x, tan x, cot x, etc. Memorise: e∫ (1/x) dx = x, e∫ (2/x) dx = x2.
Type of DEStandard FormSolution MethodExample
Variable Separabledy/dx = f(x)g(y)∫ dy/g(y) = ∫ f(x) dxdy/dx = xy
Homogeneousdy/dx = F(y/x)y = vx, separate variables(x+y) dx = x dy
Linear (first order)dy/dx + Py = QIF = e∫ P dxdy/dx + y = x
Bernoullidy/dx + Py = QynSubstitute v = y1−ndy/dx + y = xy2
ExactM dx + N dy = 0∂M/∂y = ∂N/∂x(y)dx + (x)dy = 0
Example 32 | NDA 2019
Solve the linear DE: dy/dx + y tan x = sec x
Solution: P = tan x, Q = sec x. IF = e∫ tan x dx = eln|sec x| = sec x. Solution: y sec x = ∫ sec x × sec x dx = ∫ sec2 x dx = tan x + C. So y = cos x (tan x + C) = sin x + C cos x.
⚡ NDA Shortcut: Quick DE Identification
In the NDA exam, quickly identify the type of DE: if variables separate easily, it is variable separable; if dy/dx = f(y/x), it is homogeneous; if dy/dx + Py = Q, it is linear. This saves precious time.

Bernoulli's Differential Equation

A Bernoulli DE has the form dy/dx + P(x)y = Q(x)yn where n ≠ 0, 1. To solve, substitute v = y1−n. This reduces it to a linear DE in v.

Example 33 | NDA Practice
Solve: dy/dx + y = xy2
Solution: Here n = 2. Substitute v = y1−2 = y−1. Then dv/dx = −y−2 dy/dx. Multiply the original DE by −y−2: −y−2 dy/dx − y−1 = −x. This gives dv/dx − v = −x. This is linear in v. P = −1, Q = −x. IF = e∫ −1 dx = e−x. Solution: v e−x = ∫ −x e−x dx = x e−x + e−x + C. So v = x + 1 + Cex. Substituting back: 1/y = x + 1 + Cex. Hence y = 1/(x + 1 + Cex).

Exact Differential Equations

A DE of the form M(x,y) dx + N(x,y) dy = 0 is exact if ∂M/∂y = ∂N/∂x. The solution is ∫ M dx + ∫ (terms of N not containing x) dy = C.

Example 34 | NDA Practice
Solve: (2xy + y) dx + (x2 + x) dy = 0
Solution: M = 2xy + y, N = x2 + x. ∂M/∂y = 2x + 1, ∂N/∂x = 2x + 1. Since they are equal, the DE is exact. Solution: ∫ M dx = ∫ (2xy + y) dx = x2y + xy + f(y). Then ∂/∂y of this = x2 + x + f'(y) = N = x2 + x. So f'(y) = 0 ⇒ f(y) = C. Hence x2y + xy = C.
⚡ NDA Shortcut: Exact DE Check
To quickly check if a DE is exact, compute ∂M/∂y and ∂N/∂x. If equal, it is exact. The solution is simply ∫ M dx + ∫ (N − ∂/∂y ∫ M dx) dy = C.

Applications of Differential Equations

Differential equations model real-world phenomena including growth, decay, cooling, and motion. In NDA exams, application problems often involve Newton's Law of Cooling, population growth, and radioactive decay.

Example 35 | NDA 2020
The rate of decay of a radioactive substance is proportional to its mass. If half-life is 10 years, find the fraction remaining after 20 years.
Solution: Let dm/dt = −km. Then dm/m = −k dt. Integrating: ln m = −kt + C. m = m0 e−kt. Half-life means m = m0/2 when t = 10. So 1/2 = e−10k ⇒ k = (ln 2)/10. After 20 years: m = m0 e−20(ln 2)/10 = m0 e−2 ln 2 = m0/4. So 1/4 of the original mass remains.

Practice Questions

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