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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.

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Concept 1 of 1: Solving for coefficients from integral conditions

If a function is a combination of known pieces with unknown coefficients (like Pex+Qe2x+Re3xPe^x+Qe^{2x}+Re^{3x}), then every integral, value, or derivative you are told gives one equation linking P, Q, R. Collect as many equations as unknowns and solve.

Definition

The procedure:

  • Write the unknown function with its parameters, e.g. f(x)=Pex+Qe2x+Re3xf(x)=Pe^x+Qe^{2x}+Re^{3x}.
  • Convert each condition into an equation: a value f(0)f(0), an integral ∫0cf\int_0^c f, or a derivative f′(0)=P+2Q+3Rf'(0)=P+2Q+3R.
  • 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 3f(x)+f ⁣(1x)=g(x)3f(x)+f\!\big(\tfrac1x\big)=g(x) holds for every xx, so replace xx by 1x\tfrac1x to get a second relation 3f ⁣(1x)+f(x)=g ⁣(1x)3f\!\big(\tfrac1x\big)+f(x)=g\!\big(\tfrac1x\big); solve the pair as simultaneous equations in the two unknowns f(x)f(x) and f(1/x)f(1/x) (here 8f(x)=3g(x)−g(1/x)8f(x)=3g(x)-g(1/x)). The same pairing works for x→1−xx\to1-x, x→−xx\to-x and x→a−xx\to a-x — whichever substitution maps the given arguments onto each other. Then integrate the recovered ff.

Worked example

Let f(x)=Aex+Be2xf(x)=Ae^x+Be^{2x} with f(0)=3f(0)=3 and ∫0ln⁡2f(x) dx=72\int_0^{\ln 2} f(x)\,dx = \tfrac72. Find A and B.
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The same idea in a real exam question:

NDA · Apr 2023 · Q73Moderate

Example 1 · Definite Integration · Definite Integrals in Function Conditions

Consider the following for the items that follow: Let f(x)=Pex+Qe2x+Re3xf(x)=Pe^{x}+Qe^{2x}+Re^{3x}, where PP, QQ, RR are real numbers. Further f(0)=6f(0)=6, f′(ln⁡3)=282f'(\ln3)=282 and ∫0ln⁡2f(x) dx=11\int_{0}^{\ln2}f(x)\,dx=11.
What is f′(0)f'(0) equal to?

One condition, one equation — match the counts

You need as many independent conditions as unknown parameters. With three unknowns P, Q, R you must extract three equations (e.g. a value, an integral, and a derivative) before the system is solvable.

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Test yourself on Definite Integration

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