Math 2310H , Fall 2002, Solutions to first exam 1. (15 points) This problem has two parts. The first part is worth five points. The second part is worth ten points. (a) Compute the integral Z .=2 0 sin3(x) cos( x)dx (b) Compute the integral Z . 0 cos 4(x)dx To compute the first integral, we mak e a simple substitution: u = sin( x);du = cos( x)dx: Then the integral transforms into R 1 0 u 3 du = u4 4 ] 1 0 = 1=4: To compute the second integral, we use the subtitution cos 2(x) = 1+cos(2 x)2 : Then we have Z . 0 1 + cos(2 x) 2 ¶2 dx = 14 Z . 0 ¡ 1 + 2 cos(2 x) + cos 2(2x) ¢ dx = 14 Z . 0 1 + 2 cos(2 x) + 1 + cos(4 x)2 ¶ dx = 14 x + sin(2 x) + x2 + sin(4 x)8 ¶,. 0 whic h is 3.2 or something like that. 2. (15 points) This problem has two parts. The first part is worth five points. The second part is worth ten points. (a) Compute the deriv ativ e d dx Z x 3 sin( p t + 2) dt (b) Compute the deriv ativ e d dx Z cos x 3 sin( p t + 2) dt 1 2 The firs...