Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET202017 Sep 2020Evening ShiftMathematicsDefinite IntegrationActual

( ₀^1 8 (1+x) 1+x^2 d x= )

Options

  1. A( 2 )
  2. B( 2 2 )
  3. C( 4 2 )
  4. D( 2 )

Correct answer

A. ( 2 )

Step-by-step solution

(I= ₀^1 8 (1+x) 1+x^2 d x=8 ₀^1 (1+x) 1+x^2 d x ) Let (x= d x= ^2 d ) When (x=0, =0 ) and when (x=1, = 4 ) So, ( aligned I & =8 ₀^ / 4 (1+ ) 1+ ^2 ^2 d & =8 ₀^ (1+ ) d aligned ) Now using (I= ₀^a f(x) d x= ₀^a f(a-x) d x ) and adding both equations we have, ( aligned 2 I & =8 ₀^ 2 d x I & =8 8 2= 2 aligned )

Practice Definite Integration on Quantrex Academy →

More from Definite Integration

₀^1 2 x+5 x^2+3 x+2 ~d x= 2025₀^1 x^ 5 / 2 (1-x)^ 3 / 2 ~d x= 2025_ n [ 1 n^2 ^2 1 n^2 + 2 n^2 ^2 4 n^2 + 3 n^2 ^2 9 n^2 + + 1 n^2 ^2 1 ]= 2025₀^1 x Sin ⁻¹ x d x= 2025_ - 2 ^ 2 (x-[x]) d x= 2025₀^2 x^2(2-x)^5 d x= 2025If f(x)= Max x^3-4, x^4-4 , and g(x)= Min x^2, x^3 , then _ -1 ^1(f(x)-g(x)) d x= 2025_ n 2 n [ 2 n + 2 2 n + 3 2 n + + 2 ]= 2025 Full Definite Integration list All AP EAMCET PYQs