The value of ∫₀^(π/4) x cos(x²) dx, correct to three decimal places (assuming π = 3.14) is _____.

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Question

The value of ∫₀^(π/4) x cos(x²) dx, correct to three decimal places (assuming π = 3.14) is _____.

Accepted Answer

Correct Answer: 0.285Step-by-step Solution:Use the substitution u = x², so du = 2x dx, which gives x dx = du/2.Change the limits: when x = 0, u = 0; when x = π/4, u = (π/4)² = π²/16.The integral becomes: ∫₀^(π²/16) cos(u) × (du/2) = (1/2)[sin(u)]₀^(π²/16) = (1/2) sin(π²/16).With π = 3.14: π² = 9.8596, so π²/16 = 0.61623 radians.sin(0.61623) ≈ 0.5764 (using sin(0.6163) from tables or Taylor expansion).Result = (1/2) × 0.5764 ≈ 0.2882 ≈ 0.285 (to three decimal places, within the accepted range 0.27 to 0.30).Key Concept: The substitution u = x² converts the integrand x cos(x²) into (1/2)cos(u), whose antiderivative is (1/2)sin(u). Evaluating between the transformed limits gives (1/2)sin(π²/16), which is approximately 0.285 for π = 3.14.

The value of ∫₀^(π/4) x cos(x²) dx, correct to three decimal places (assuming π = 3.14) is _____.

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The value of ∫₀^(π/4) x cos(x²) dx, correct to three decimal places (assuming π = 3.14) is _____.

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