If $\int_0^xf(t)dt=x+\int_x^1t\,f(t)dt$ then the value of f(1) is:
Answer & explanation
Correct answer: option 1
Differentiate w.r.t. x ⇒ f(x) – 0 = 1 + 0 – x f(x) × 1 ; Put x = 1 ⇒ f(1) = 1 – f(1) ⇒ f(1) = $\frac{1}{2}$
If $\int_0^xf(t)dt=x+\int_x^1t\,f(t)dt$ then the value of f(1) is:
Correct answer: option 1
Differentiate w.r.t. x ⇒ f(x) – 0 = 1 + 0 – x f(x) × 1 ; Put x = 1 ⇒ f(1) = 1 – f(1) ⇒ f(1) = $\frac{1}{2}$