Definite IntegrationHard
Question
Let f be a non-negative function defined on the interval [0, 1]. If 
f(t)dt, 0 ≤ x ≤ 1, and f(0) = 0, then
f(t)dt, 0 ≤ x ≤ 1, and f(0) = 0, then Options
A.
B.
C.
D.
Solution
f′ = ±
⇒ f(x) = sin x or f(x) = - sin x (not possible)
⇒ f(x) = sin x
Also, x > sin x ∀ x > 0.
⇒ f(x) = sin x or f(x) = - sin x (not possible)
⇒ f(x) = sin x
Also, x > sin x ∀ x > 0.
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