Question

$If \int\limits^k_2 (2x+1) dx=6, then k=.....,Select correct option from the given options.$

(a) 3

(b) 4

(c) -4

(d) -2

Answer 1

HELLO DEAR,

$\bold{\int\limits^k_2 (2x+1) dx= 6}$

$\bold{[2x^2/2 + x]^{^k}_2 = 6}$

$\bold{[(k)^2 + k - (2)^2 - 2] = 6}$

$\bold{k^2 + k - 6 = 6}$

$\bold{k^2 + k - 12 = 0}$

$\bold{k^2 + 4k - 3k - 12 = 0}$

$\bold{k(k + 4) - 3(k + 4) = 0}$

$\bold{(k - 3)(k + 4) = 0}$

k = -4 , k = 3
but k ≠ -4 because upper limit will be negative. then, integration will negative.


hence, option (a) is correct,

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int\limits^k_2{(2x+1)}\,dx$ = 6

$\int\limits^k_2{(2x+1)}\,dx$ = 6

$\implies\int\limits^k_2{2x}\,dx+\int\limits^k_2dx=6$

$\implies[x^2]^k_2+[x]^k_2=6$

$\implies(k^2-2^2)+(k-2)=6$

$\implies(k^2+k-6-6)=0$

k² + k - 12 = 0

k² + 4k - 3k - 12 = 0

k(k + 4) - 3(k + 4) = 0

(k + 4)(k - 3) = 0

k = -4 , 3

but k ≠ -4 because upper limit will be negative. then, integration will negative. but we see , integration is positive e.g., 6

hence, k = 3
option (a) is correct.