Let [ t ] denote the largest integer less than or equal to t . If 0 3 ( [ x 2 ] + [ x 2 2 ] ) d x = a + b 2 - 3 - 5 + c 6 - 7 , where a , b , c Z , then a + b + c 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 $[t]$ denote the largest integer less than or equal to $t .$ If $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7},$ where $a, b, c \in Z,$ then $a + b + c$ is equal to _______. [2024]

Ans.
(23)

Given, $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7}$

Now, $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\sqrt{3}} (2 + 1) d x$

$+ \int_{\sqrt{3}}^{2} (3 + 1) d x + \int_{2}^{\sqrt{5}} 6 d x + \int_{\sqrt{5}}^{\sqrt{6}} 7 d x + \int_{\sqrt{6}}^{\sqrt{7}} 9 d x + \int_{\sqrt{7}}^{\sqrt{8}} 10 d x + \int_{\sqrt{8}}^{3} 12 d x$

$= (\sqrt{2} - 1) + (3 \sqrt{3} - 3 \sqrt{2}) + (8 - 4 \sqrt{3}) + (6 \sqrt{5} - 12) + (7 \sqrt{6} - 7 \sqrt{5}) + (9 \sqrt{7} - 9 \sqrt{6}) + (10 \sqrt{8} - 10 \sqrt{7}) + (36 - 12 \sqrt{8})$

$= 31 - 6 \sqrt{2} - \sqrt{3} - \sqrt{5} - 2 \sqrt{6} - \sqrt{7}$

$\Rightarrow a = 31, b = - 6, c = - 2$

So, $a + b + c = 31 - 8 = 23$

Want Us to Email you About Special Offers & Updates?