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Integrals
Previous Year Questions (2022-2024)
Year 2022
- Evaluate ∫dxx2 + 4.
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∫dxx2 + 4 = 12tan^{-1}(x2) + C - Evaluate ∫_0^π/2 sin x dx.
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∫_0^π/2 sin x dx = [-cos x]_0^π/2 = 0 - (-1) = 1
Year 2023
- Evaluate ∫dxx2 + 4.
Show Solution
∫dxx2 + 4 = 12tan^{-1}(x2) + C - Evaluate ∫_0^π/2 sin x dx.
Show Solution
∫_0^π/2 sin x dx = [-cos x]_0^π/2 = 0 - (-1) = 1
Year 2024
- Evaluate ∫dxx2 + 4.
Show Solution
∫dxx2 + 4 = 12tan^{-1}(x2) + C - Evaluate ∫_0^π/2 sin x dx.
Show Solution
∫_0^π/2 sin x dx = [-cos x]_0^π/2 = 0 - (-1) = 1