Let \(f:(0, \infty) \rightarrow R\) and \(F(x)=\int_{0}^{x} t f(t) d t\). If \(F(x^{2})= x^{4}+x^{5}\), then \(\sum_{r=1}^{12} f(r^{2})\) is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Definite Integrals

Let

f : (0, \infty) \to R

and

F (x) = \int_{0}^{x} t f (t) d t

. If

F (x^{2}) = x^{4} + x^{5}

, then

\sum_{r = 1}^{12} f (r^{2})

is equal to :

Answer: 219

F (x) = \int_{0}^{x} t f (t) d t

, so by the Fundamental Theorem of Calculus,

F^{'} (u) = u f (u)

.

Given

F (x^{2}) = x^{4} + x^{5}

. Differentiate both sides with respect to

x

, using the chain rule on the left:

\frac{d}{d x} F (x^{2}) = F^{'} (x^{2}) \cdot 2 x = [x^{2} f (x^{2})] \cdot 2 x = 2 x^{3} f (x^{2}) .

Right side:

\frac{d}{d x} (x^{4} + x^{5}) = 4 x^{3} + 5 x^{4}

2 x^{3} f (x^{2}) = 4 x^{3} + 5 x^{4} ⟹ f (x^{2}) = 2 + \frac{5}{2} x .

Setting

x = r

(positive, since the domain is

(0, \infty)

f (r^{2}) = 2 + \frac{5}{2} r

r = 1 \sum 12 f (r^{2}) = r = 1 \sum 12 (2 + \frac{5}{2} r) = 24 + \frac{5}{2} \cdot \frac{12 \cdot 13}{2} = 24 + \frac{5}{2} (78) = 24 + 195 = 219.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer