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

Intuition

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.

Worked example

Let f(x)=Aex+Be2xf(x)=Ae^x+Be^{2x} with f(0)=3f(0)=3 and 0ln2f(x)dx=72\int_0^{\ln 2} f(x)\,dx = \tfrac72. Find A and B.
  1. f(0)=A+B=3f(0)=A+B=3.
  2. 0ln2f=[Aex+B2e2x]0ln2=(2A+2B)(A+B2)=A+32B=72\int_0^{\ln2} f = [Ae^x + \tfrac{B}{2}e^{2x}]_0^{\ln2} = (2A+2B)-(A+\tfrac{B}{2}) = A + \tfrac32 B = \tfrac72.
  3. Solve A+B=3, A+32B=72A+B=3,\ A+\tfrac32 B=\tfrac72: subtract to get 12B=12B=1, A=2\tfrac12 B=\tfrac12\Rightarrow B=1,\ A=2.
Answer:A=2, B=1A=2,\ B=1.
Practice this conceptself-check · 3 quick reps

Try it yourself

For f(x)=Pex+Qe2x+Re3xf(x)=Pe^x+Qe^{2x}+Re^{3x} it is found that P=1,Q=2,R=3P=1,Q=2,R=3. What is f(0)f'(0)?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    How many independent conditions to fix 3 unknown coefficients?
  2. 2.
    For f=Pex+Qe2x+Re3xf=Pe^x+Qe^{2x}+Re^{3x}, write f(0)f'(0).
  3. 3.
    0ce2xdx=?\int_0^{c} e^{2x}\,dx = ?

From the bank · past-year question

Example 1Definite IntegrationMODERATE
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(ln3)=282f'(\ln3)=282 and 0ln2f(x)dx=11\int_{0}^{\ln2}f(x)\,dx=11.
What is f(0)f'(0) equal to?

[Q73 · Apr · 2023]

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.

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)

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Definite IntegrationHARD
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(ln3)=282f'(\ln3)=282 and 0ln2f(x)dx=11\int_{0}^{\ln2}f(x)\,dx=11.
What is the value of RR?

[Q72 · Apr · 2023]

Example 2Definite IntegrationHARD
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(ln3)=282f'(\ln3)=282 and 0ln2f(x)dx=11\int_{0}^{\ln2}f(x)\,dx=11.
What is the value of QQ?

[Q71 · Apr · 2023]

Drill every past-year question on this subtopic

3 questions from the bank — paginated, with cart and Word-export support.