If _0^K 0.05em dx2+18x^2= 24, then the value of K is

Question

Accepted Answer

Correct answer: C. (c) : Let I= _0^K 0.05em dx2+18x^2= 118 _0^K 0.05em dx 218+x^2 ⇒I= 118 _0^K 0.05em dx 19+x^2= 118 _0^K 0.05em dx ( 13 )^2+x^2 ⇒I= 118 ( 1 ( 13 ) tan^-1 ⁡ ( x 13 ) )_0^K= 24 ⇒ tan^-1 ⁡3K= 4= tan^-1 ⁡1 ⇒3K=1 ⇒K= 13.

$If \int_{0}^{K} \frac{dx}{2 + 18 x ^{2}} = \frac{π}{24}, then the value of K is$

Solution

More Definite Integration PYQs