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JEE Main20266 April 2026Evening ShiftMathematicsDefinite IntegrationActual

The value of the integral _ -1 ¹ ( x^3 + |x| + 1 x^2 + 2|x| + 1 ) dx is equal to :

Options

  1. A3 _e 2
  2. B2 _e 2
  3. C5 _e 3
  4. D3 _e 3

Correct answer

B. 2 _e 2

Step-by-step solution

Let I = _ -1 ¹ ( x^3 + |x| + 1 x^2 + 2|x| + 1 ) dx We can split the integral into two parts: I = _ -1 ¹ x^3 x^2 + 2|x| + 1 dx + _ -1 ¹ |x| + 1 x^2 + 2|x| + 1 dx The first integrand f(x) = x^3 x^2 + 2|x| + 1 is an odd function since f(-x) = -f(x) . Therefore, its integral over the symmetric interval [-1, 1] is zero. The second integrand g(x) = |x| + 1 x^2 + 2|x| + 1 is an even function since g(-x) = g(x) . Therefore, its integral over [-1, 1] is twice the integral over [0, 1] . I = 0 + 2 ₀¹ x + 1 x^2 + 2x + 1 dx I =

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