MATH116 Lecture Notes - Lecture 1: Even And Odd Functions

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Tutorial solutions 10: evaluate the following de nite integrals (cid:90) 4. 0 (1 + sin t) dt (c) (cid:90) a. A (c) es e s ds (cid:90) 2 . 1 + sin t dt = t cos t. = 2 cos 2 (0 cos 0) |x 2| dx = (cid:90) 2 (cid:90) 2 (cid:90) 2. (x 2) dx + x 2 dx (cid:12)(cid:12)(cid:12)(cid:12)2 x 2 dx + x 2 dx. = (2 4) + (8 8 (2 4)) 2 (cid:90) (cid:90: compute the following integrals. A suitable substitution (possibly after some manipula- tion of the integrand) should work in each case. x2 + 1 dx x3 (b) (a) (cid:90) . 0 (cos2(x) sin(x) + 3 sin(x)) dx (c) x sin(x2 + 2) dx. Solution: (a) let u = x2 + 1, then du = 2x dx. Note that x3 can be broken up as x x2. In particular, we can replace the x dx with 1.

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