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Manipal MET2010MathematicsDefinite Integration

The approximate value of ₁^5 x^2 d x using trapezoidal rule with n=4 is

Options

  1. A41
  2. B41.5
  3. C41.75
  4. D42

Correct answer

D. 42

Step-by-step solution

₁^5 x^2 d x= 4 4 [ 1 2 f(1)+f(2)+f(3)+f(4)+ 1 2 f(5) ]= [ 1 2 +4+9+16+ 25 2 ]=42 [ . using .f(x)=x^2 ]

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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 Manipal MET PYQs