JEE MainMathematicsDefinite Integration
The value of the integral _ - 4 ^ 4 dx (1 + ^2 x)(1 + 2024^x) is equal to
Options
- A⁻¹ ( 1 2 )
- B1 2 ⁻¹ ( 1 2 )
- C2 ⁻¹ ( 1 2 )
- D1 2 ⁻¹ ( 4 2 )
Correct answer
B. 1 2 ⁻¹ ( 1 2 )
Step-by-step solution
Let I = _ - /4 ^ /4 dx (1 + ^2 x)(1 + 2024^x) Using the property _ -a ^a f(x) dx = ₀^a (f(x) + f(-x)) dx , we have: I = ₀^ /4 ( 1 (1 + ^2 x)(1 + 2024^x) + 1 (1 + ^2(-x))(1 + 2024^ -x ) ) dx I = ₀^ /4 1 1 + ^2 x ( 1 1 + 2024^x + 2024^x 2024^x + 1 ) dx I = ₀^ /4 1 1 + ^2 x dx Dividing the numerator and the denominator by ^2 x : I = ₀^ /4 ^2 x ^2 x + 1 dx = ₀^ /4 ^2 x 2 + ^2 x dx Let x = t ^2 x dx = dt . When x = 0, t = 0 and when x = 4 , t = 1 . I = ₀^1 dt ( 2 )^2 + t^2 I = 1 2 [ ⁻¹ ( t 2 ) ]₀^1 = 1 2 ⁻¹ ( 1 2 ) Answ