Solveeit Logo
Login

Question

Mathematics Question on Definite Integral

Let f be a real valued continuous function on [0, 1] and
f(x)=x+∫01(x−t)f(t) dtf(x) = x + \int_{0}^{1} (x - t) f(t) \,dt
Then, which of the following points (x, y) lies on the curve y = f(x)?

A

(2, 4)

B

(1, 2)

C

(4, 17)

D

(6, 8)

Answer

(6, 8)

Explanation

Solution

The correct answer is (D) : (6,8)
f(x)=x∫01(x−t)f(t) dtf(x) = x \int_{0}^{1} (x - t) f(t) \,dt
f(x)=x+x∫01f(t) dt−∫01t⋅f(t) dtf(x) = x + x\int_{0}^{1} f(t) \,dt - \int_{0}^{1} t \cdot f(t) \,dt
f(x)=x(1+∫01f(t) dt)−∫01t⋅f(t) dtf(x) = x \left(1 + \int_{0}^{1} f(t) \,dt \right) - \int_{0}^{1} t \cdot f(t) \,dt
Let
1+∫01f(t) dt=aand∫01t⋅f(t) dt=11 + \int_{0}^{1} f(t) \,dt = a \quad \text{and} \quad \int_{0}^{1} t \cdot f(t) \,dt = 1
f(x) = ax-b
Now,
a=1+∫01(at−b) dt=1+a2−b  ⟹  a2+b=1a = 1 + \int_{0}^{1} (at - b) \,dt = 1 + \frac{a}{2} - b \implies \frac{a}{2} + b = 1
b=∫01t(at−b) dt=a3−b2  ⟹  3b2=a3  ⟹  b=2a9b = \int_{0}^{1} t(at - b) \,dt = \frac{a}{3} - \frac{b}{2} \implies \frac{3b}{2} = \frac{a}{3} \implies b = \frac{2a}{9}
a2+2a9=1\frac{a}{2} + \frac{2a}{9} = 1
⇒a=1813⇒ a = \frac{18}{13} b=413b = \frac{4}{13}
ƒ(x)=18x−413ƒ(x) = \frac{18x-4}{13}
(6,8) lies on f(x)