NDA Maths · Definite Integration
Recovering a Function from Integral Conditions
When an unknown function has parameters and you are given several integral or derivative conditions, each condition becomes one linear equation — solve the system for the parameters.
Why this matters
A small (3 PYQ) but reliably HARD subtopic. The work is bookkeeping: turn each given condition (a definite integral, a value, a derivative) into an equation in the unknown coefficients, then solve the linear system.
Concept 1 of 1: Solving for coefficients from integral conditions
Definition
The procedure:
- Write the unknown function with its parameters, e.g. .
- Convert each condition into an equation: a value , an integral , or a derivative .
- Solve the resulting linear system for the parameters, then answer the specific question asked.
When the unknown is the FUNCTION, not its coefficients — the reciprocal-argument trick: a relation such as holds for every , so replace by to get a second relation ; solve the pair as simultaneous equations in the two unknowns and (here ). The same pairing works for , and — whichever substitution maps the given arguments onto each other. Then integrate the recovered .
Worked example
Practice this conceptself-check · 3 quick reps
The same idea in a real exam question:
Example 1 · Definite Integration · Definite Integrals in Function Conditions
One condition, one equation — match the counts
Summary — formulas & gotchas at a glance
A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.
Watch out for (1)
- One condition, one equation — match the counts→ Solving for coefficients from integral conditions
Test yourself on Definite Integration
15 past NDA questions from this chapter, timed at 19 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.