Let f : ( 0 , )R and F ( x ) = 0 x t f ( t ) ? d t . If F ( x 2 ) = x 4 + x 5 , then r = 1 12 f ( r 2 ) is equal to _ _ _ _ . [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
$Let f : (0, \infty) \to R and F (x) = \int_{0}^{x} t f (t) d t . If F (x^{2}) = x^{4} + x^{5}, then \sum_{r = 1}^{12} f (r^{2}) is equal to____.$ [2024]

Ans.
(219)

We have, $F (x) = \int_{0}^{x} t \cdot f (t) d t$

$\Rightarrow F' (x) = x \cdot f (x)$ ...(i)

Also, we have $F (x^{2}) = x^{4} + x^{5}$

Differentiating on both sides, we get

$\Rightarrow 2 x \cdot F' (x^{2}) = 4 x^{3} + 5 x^{4}$

$\Rightarrow F' (x^{2}) = 2 x^{2} + \frac{5}{2} x^{3} \Rightarrow F' (x) = 2 x + \frac{5}{2} x^{3 / 2}$ ...(ii)

From (i) & (ii), we get $\Rightarrow x f (x) = 2 x + \frac{5}{2} x^{3 / 2}$

$\Rightarrow f (x) = 2 + \frac{5}{2} x^{1 / 2} \Rightarrow f (x^{2}) = 2 + \frac{5}{2} x$

$∴ \sum_{r = 1}^{12} f (r^{2}) = \sum_{r = 1}^{12} (2 + \frac{5}{2} r) = 24 + \frac{5}{2} \cdot \frac{12 \cdot 13}{2} = 219$

Want Us to Email you About Special Offers & Updates?