Chapter 2. Methods of Integration
1 x2 5 p x dx. Solution: By the Fundamental Theorem of Calculus and Linearity, we have Z 4 1 x2 5 p x dx= Z 4 1 x3 = 255x 1 dx= h 5 x =2 10x1=2 i 4 1 = 64 20 10 = 12: 2.3 Example: Find Zp ˇ=3 ˇ=6 sin2x+ cos3xdx. Solution: We nd antiderivatives for sin2xand cos3x. Since d dx cos2x= 2sin2xwe have d dx 1 2 cos2x = sin2xand since d dx sin3x ... ................
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