009B Sample Midterm 3, Problem 3 Detailed Solution

Find a curve  $y=f(x)$  with the following properties:

(i)   $f''(x)=6x$

(ii)   Its graph passes through the point  $(0,1)$  and has a horizontal tangent there.

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Background Information:
1. If the graph of  $f(x)$  passes through the point  $(a,b),$  then  $f(a)=b.$
2. If  $f(x)$  has a horizontal tangent at the point  $(a,b),$  then  $f'(a)=0.$

Solution:

Step 1:
Let  $f(x)=\sin(x).$
Each interval has length  ${\frac {\pi }{4}}.$
Therefore, the right-endpoint Riemann sum of  $f(x)$  on the interval  $[0,\pi ]$  is
${\frac {\pi }{4}}{\bigg (}f{\bigg (}{\frac {\pi }{4}}{\bigg )}+f{\bigg (}{\frac {\pi }{2}}{\bigg )}+f{\bigg (}{\frac {3\pi }{4}}{\bigg )}+f(\pi ){\bigg )}.$

Step 2:
Thus, the right-endpoint Riemann sum is
${\begin{array}{rcl}\displaystyle {{\frac {\pi }{4}}{\bigg (}\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}+\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}+\sin {\bigg (}{\frac {3\pi }{4}}{\bigg )}+\sin(\pi ){\bigg )}}&=&\displaystyle {{\frac {\pi }{4}}{\bigg (}{\frac {\sqrt {2}}{2}}+1+{\frac {\sqrt {2}}{2}}+0{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{4}}({\sqrt {2}}+1).}\\\end{array}}$

Final Answer:
${\frac {\pi }{4}}({\sqrt {2}}+1)$

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