JEE Advanced 2020 Paper 1 · Mathematics · Indefinite and Definite Integrals · medium

A function f:mathbb R→mathbb R satisfies f(x)=x^2+2x∫0^1t f(t) dt for all real x. Find ∫0^2f(x) dx.

Numerical answer

A function

f:\mathbb R\to\mathbb R

satisfies

f(x)=x^{2}+2x\displaystyle\int_{0}^{1}t\,f(t)\,dt

for all real

x

. Find

\displaystyle\int_{0}^{2}f(x)\,dx

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 5.67

The definite integral

k=\int_0^1 t f(t)\,dt

is a constant, so

f(x)=x^2+2kx

. Substituting back,

k=\int_0^1 t(t^2+2kt)\,dt=\dfrac14+\dfrac{2k}{3}

. Solving,

k\left(1-\dfrac23\right)=\dfrac14

, so

k=\dfrac34

. Therefore

f(x)=x^2+\dfrac32x

. Now

\int_0^2f(x)\,dx=\left[\dfrac{x^3}{3}+\dfrac{3x^2}{4}\right]_0^2=\dfrac83+3=\dfrac{17}{3}\approx5.67

. A symbolic solve of the same equations gives the identical coefficient

3/2

and the same integral. The self-consistency step on

k

is the key idea; evaluating

f

before determining

k

would leave an unknown constant.

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