M403L First Exam Solutions

September 15, 1998

Problem 1: Let tex2html_wrap_inline65 . Find f'(x).

By the product rule combined with the chain rule, tex2html_wrap_inline69 .

Problem 2: Let tex2html_wrap_inline71 . Find dy/dx.

By the ratio rule, tex2html_wrap_inline75 .

Problem 3: Evaluate tex2html_wrap_inline77

Substitute tex2html_wrap_inline79 . tex2html_wrap_inline81 .

Problem 4: Find tex2html_wrap_inline83

Integrate by parts. tex2html_wrap_inline85 , dU=dx/x, tex2html_wrap_inline89 , tex2html_wrap_inline91 , so tex2html_wrap_inline93 .

Problem 5: Evaluate tex2html_wrap_inline95

Integrate by parts. U=x, dU=dx, tex2html_wrap_inline101 , tex2html_wrap_inline103 , so tex2html_wrap_inline105 . The limit as tex2html_wrap_inline107 is zero, while evaluating at x=0 gives -4, so the definite integral is 0-(-4)=4.

Problem 6: Find tex2html_wrap_inline115 .

Substitute u=x+1: tex2html_wrap_inline119 . Alternatively, you can integrate by parts to get tex2html_wrap_inline121 . These answers look different, but they're actually the same since tex2html_wrap_inline123 .

Problem 7: Evaluate tex2html_wrap_inline125

Substitute tex2html_wrap_inline127 , du=dx/x. tex2html_wrap_inline131 . Plugging in at x=1 and x=e gives 1/4 - 0/4 = 1/4.

Problem 8: Let tex2html_wrap_inline139 . Find tex2html_wrap_inline141 and tex2html_wrap_inline143 .

tex2html_wrap_inline145 ; tex2html_wrap_inline147 .

Problem 9: Let tex2html_wrap_inline139 (same as problem 8). Find tex2html_wrap_inline151 , tex2html_wrap_inline153 and tex2html_wrap_inline155 .

tex2html_wrap_inline157 ; tex2html_wrap_inline159 ; tex2html_wrap_inline161 .

Problem 10: Let tex2html_wrap_inline163 . Where does f(x,y) achieve its minimum value? What is that minimum value?

Solve tex2html_wrap_inline167 ; tex2html_wrap_inline169 . The solution is x = -1, y=1. The minimum value is f(-1,1)=4.


Lorenzo Sadun
Wed Sep 16 17:13:02 CDT 1998