Fundamentals
The core facts every aspirant should own — each a titled nugget with a real-world story, the concept in plain words, and a memory trick. Works even when the internet doesn't.
1,200 fundamentals
“Measuring water flowing between two bridges is a definite integral.”
The definite integral ∫_aᵇ f(x)dx equals the signed area between the curve and the x-axis from a to b. Area above the axis is positive, below is negative, and swapping limits reverses the sign. For total (unsigned) area split the interval at the zeros of f(x).
Memory trick: Upper-lower → sign; swap limits → opposite sign.
“A cyclist’s net displacement counts downhill as negative, just like a signed integral.”
A frequent slip is to assume the integral always gives a positive area. The definite integral actually yields signed area, so portions of the curve below the x-axis subtract from the total. To get the total (unsigned) area, break the interval at each root and add the absolute values.
Memory trick: Below axis = negative; add |area| for total.
“When you see ∫f(x)·f'(x)dx, it's like a lock and key—substitution opens it instantly.”
If the integrand contains a function and its exact derivative, let u = that function. Then du = f'(x)dx and the integral becomes ∫u du, which is trivial. The trap: forcing substitution when the derivative isn’t present wastes time; switch to parts or other tricks then.
Memory trick: f·f' → SUB first
“When navigating a campus, you first look for the straight path before wandering through the garden, just like trying substitution before by-parts.”
A simple u-substitution often turns the integral into a basic form; by-parts adds extra terms and can complicate it. The exam trap is using by-parts on ∫f(g(x))g'(x)dx where u = g(x) works instantly. Always check if the integrand contains a function and its derivative first.
Memory trick: Derivative of inner present? → try SUB first
“Climbing stairs where each step height follows xⁿ gives total height x^(n+1)/(n+1).”
The power rule ∫xⁿdx = xⁿ⁺¹/(n+1)+C holds for any real n except n = –1. It follows from reversing the differentiation of xⁿ⁺¹. Remember that n = –1 leads to the logarithmic case, not division by zero.
∫xⁿ dx = x^(n+1)/(n+1) + C
Memory trick: n≠-1 → add 1, divide.
“Continuous 100% interest growth over time integrates to ln|x|.”
∫(1/x)dx = ln|x|+C for x≠0. The absolute value ensures the antiderivative works for both positive and negative x because d/dx[ln|x|]=1/x. Omitting the | | gives the wrong sign on the negative side.
∫(1/x) dx = ln|x| + C
Memory trick: 1/x → ln|x|, keep absolute.
“When you measure water collected in a bucket you get one volume; writing the formula for how fast the tap fills gives a family of curves.”
A definite integral ∫_aᵇ f(x)dx has upper and lower limits and evaluates to a single number, the net area. An indefinite integral ∫ f(x)dx has no limits; its result is a general antiderivative F(x)+C, a family of functions. In JEE, missing the limits or adding +C to a definite integral loses marks.
Memory trick: Definite = ‘Done’ → number; Indefinite = ‘Infinite’ → family (+C).
“A speedometer shows the car’s instant speed (derivative); the odometer shows total distance travelled over time (integral).”
Differentiation turns y=f(x) into its instantaneous rate dy/dx, the slope at each point. Integration reverses this, accumulating infinitesimal pieces to recover the original function (plus C) or total area. Trap: treating ∫ f'(x)dx as f(x) without the +C.
Memory trick: Diff = slope, Int = sum – think ‘D’ for ‘Derivative’, ‘I’ for ‘Integral’ = ‘Add up’.
“Finding a particle’s position from its velocity requires adding the unknown starting point, otherwise the path is incomplete.”
Every indefinite integral ∫ f(x)dx yields an antiderivative plus an arbitrary constant C because differentiation removes constants. Omitting +C makes the answer incomplete and can cost marks, especially when later applying initial conditions. Trap: writing ∫ x dx = x²/2 without +C.
Memory trick: Indefinite Integral → Include C.
“Rolling a fair die, the chance of a 4 is 1 favourable outcome out of 6 total outcomes.”
Probability = favourable outcomes ÷ total outcomes, so 0 ≤ P ≤ 1. Always list all equally likely outcomes before dividing. If the result exceeds 1, the sample space was counted incorrectly.
Memory trick: P>1? Re-count the sample space.
“Thinking the chance of drawing a red card is 14/13 shows you counted extra cards.”
A probability greater than 1 is impossible; it means the numerator or denominator is wrong—usually double-counted favourable cases or omitted equally likely outcomes. Verify the sample space first.
Memory trick: P>1 → recount outcomes.
“Two light switches that can never be on together act like mutually exclusive events.”
Independent events satisfy P(A∩B)=P(A)P(B); mutually exclusive events have P(A∩B)=0 and are therefore dependent because one occurring stops the other. Do not treat exclusive events as independent.
Memory trick: Exclusive → zero overlap.
“Seeing a friend’s “online” badge (B) and asking the chance they’ll reply (A) uses P(A|B), not the reverse.”
P(A|B) is the probability that A occurs when B is known to have happened; mathematically P(A|B)=P(A∩B)/P(B) (P(B)>0). A common error is swapping A and B, giving P(B|A)=P(A∩B)/P(A). They match only in trivial zero-probability cases or when P(A)=P(B).
Memory trick: Condition → denominator: P(A|B)=P(A∩B) ÷ P(B).
“The chance of drawing a heart or a king equals P(heart)+P(king)−P(heart∧king).”
For any two events, P(A∪B)=P(A)+P(B)−P(A∩B). The subtraction removes the double-counted overlap. Forgetting the intersection overestimates the probability.
P(A∪B) = P(A)+P(B)−P(A∩B)
Memory trick: Add, then subtract overlap.
“In a bag, the chance of a red ball given you picked from the left half is (red∧left)/P(left).”
Conditional probability formula P(A|B)=P(A∩B)/P(B) holds only when P(B)>0; otherwise the expression is undefined. Always check the denominator first.
P(A|B) = P(A∩B)/P(B)
Memory trick: Denominator must be >0.
“Flipping two coins: each result doesn't change the other's chance (independent); picking a red card means you can't pick a black one in that same draw (mutually exclusive).”
Independent events satisfy P(A∩B)=P(A)·P(B). Mutually exclusive events have P(A∩B)=0, so P(A|B)=0 and they are dependent unless one event has zero probability. Exam trap: using the multiplication rule for exclusive events gives zero only when an event is impossible.
Memory trick: Indep → multiply; Exclusive → zero overlap.
“A doctor sees a positive test (B) and wonders about the disease chance (A); the reverse asks the chance a diseased patient tests positive.”
P(A|B)=P(A∩B)/P(B) requires P(B)>0; P(B|A)=P(A∩B)/P(A) requires P(A)>0. They differ in general. When both are defined, they are equal only if the events are disjoint (P(A∩B)=0) or their marginal probabilities are equal, P(A)=P(B).
Memory trick: Equal ⇔ same marginal or no overlap (P(A∩B)=0).
“If a dice shows 6, it cannot simultaneously show 1 – the events are mutually exclusive, not independent.”
Mutually exclusive events give P(A∩B)=0, so P(A|B)=0, showing they are dependent. Independence requires P(A∩B)=P(A)P(B). The myth fails because zero overlap equals independence only when one event’s probability is zero.
Memory trick: Zero overlap ≠ independent unless impossible.
“When you push a swing, the work you do equals the dot product of force and displacement.”
The dot product a·b = |a||b|cosθ is a scalar. It measures how much one vector projects onto the other; it becomes zero when the vectors are perpendicular. A common trap is to treat the result as a vector or to forget the cosine factor.
Memory trick: Dot = scalar; a dot is a point → one number.
“The torque you feel turning a wrench is the cross product of force and lever arm.”
The cross product a×b = |a||b|sinθ n̂ is a vector perpendicular to both a and b, with magnitude equal to the area of the parallelogram they span. Students often mistake it for a scalar or ignore the right-hand rule for its direction.
Memory trick: Cross = vector; X points out of the page (right-hand rule).
“A kite’s line direction in the sky is described by its direction ratios (dx, dy, dz).”
Direction ratios (l, m, n) are any set of numbers proportional to the components of a line’s direction vector. They define the line’s orientation in 3-D; any multiple represents the same direction. Remember they are not normalized.
Memory trick: DR are raw components; any multiple works.
“Aligning a satellite dish uses direction cosines, not raw direction ratios.”
Direction cosines (α, β, γ) are the normalized direction ratios: α = l/√(l²+m²+n²) etc., and satisfy α²+β²+γ²=1. Confusing them with raw ratios leads to wrong angle calculations.
Memory trick: Cosine = DR divided by its length.
“Two arrows pointing the same way have dot product = product of their lengths.”
a·b = |a||b|cosθ gives a scalar equal to the product of the magnitudes times the cosine of the angle between them. Use it to find the angle or projection; forgetting the absolute value signs or using degrees incorrectly is a frequent error.
a·b = |a||b|cosθ
Memory trick: Dot = |a||b|·cosθ.
“The cross product’s magnitude is the area of the parallelogram formed by the two side-vectors, like the floor you’d tile using two fence lengths at an angle.”
|a×b| = |a||b|sinθ gives the area of the parallelogram spanned by a and b. It vanishes when the vectors are collinear (θ=0 or π) or when either vector is the zero vector. In JEE, remember to include the anti-parallel and zero-vector cases.
|a×b| = |a||b|sinθ
Memory trick: Area = side·side·sinθ → zero if sinθ=0 or a=0 or b=0