Question

$If \int\limits^a_1 (3x²+2x+1)dx=11, then a=......,Select correct option from the given options.$

(a) 2

(b) 3

(c) -3

(d) 2/3

Answer 1

HELLO DEAR,


GIVEN:- $\sf{\int\limits^a_1 (3x^2+2x+1)dx=11}$

now,

$\sf{\int\limits^a_1 (3x^2.dx + 2x.dx + 1.dx) = 11}$

$\sf{ [3x^3/3 + 2x^2/2 + x]^{^a}_{_1} = 11}$

=> (a)³ + (a)² + (a) - (1 + 1 + 1)= 11

=> a³ + a² + a = 14

=> a³ + a² + a = 8 + 4 + 2

=> a³ + a² + a = (2)³ + (2)² + (2)

on comparing both side
we get,

a = 2


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Given, $\int\limits^a_1{3x^2+2x+1}\,dx=11$

=> $\int\limits^a_1{3x^2}\,dx+\int\limits^a_1{2x}\,dx+\int\limits^a_1dx=11$

=> $[x^3]^a_1+[x^2]^a_1+[x]^a_1=11$

=> $(a^3-1^2)+(a^2-1)+(a-1)=11$

=>$a^3+a^2+a-14=0$

=> a³ + a² + a - 14 = 0

=> a³ - 2a² + 3a² - 6a + 7a - 14 = 0

=> a²(a - 2) + 3a(a - 2) + 7(a - 2) = 0

=> (a² + 3a + 7)(a - 2) = 0

a² + 3a + 7 ≠ 0 , a = 2

hence, option (a) is correct.