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Given d/dx F(x) = 1/√(2x – x²) and F(1) = 0, find F(x).
If ∫₋₂³ x² dx = k∫₀² x² dx + ∫₂³ x² dx, then the value of k is :
∫₀^(π/6) sec²(x − π/6) dx is equal to :
Evaluate ∫_(log√2)^(log√3) 1/((eˣ + e⁻ˣ)(eˣ - e⁻ˣ)) dx
If ∫₀ᵃ 3x² dx = 8, then the value of 'a' is :
The value of ∫₀^(π/6) sin 3x dx is :
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