007B Sample Midterm 1, Problem 5 Detailed Solution

Find the area bounded by  $y=\sin(x)$  and  $y=\cos(x)$  from  $x=0$  to  $x={\frac {\pi }{4}}.$

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Background Information:
1. You can find the intersection points of two functions, say  $f(x),g(x),$
by setting  $f(x)=g(x)$  and solving for  $x.$
2. The area between two functions,  $f(x)$  and  $g(x),$  is given by  $\int _{a}^{b}f(x)-g(x)~dx$
for  $a\leq x\leq b,$  where  $f(x)$  is the upper function and  $g(x)$  is the lower function.

Solution:

Step 1:
Setting  $\cos x=2-\cos x,$  we get  $2\cos x=2.$
Therefore, we have
$\cos x=1.$
In the interval  $0\leq x\leq 2\pi ,$  the solutions to this equation are
$x=0$  and  $x=2\pi .$
Plugging these values into our equations,
we get the intersection points  $(0,1)$  and  $(2\pi ,1).$

Step 2:
We now have
${\begin{array}{rcl}\displaystyle {\int x^{2}{\sqrt {1+x^{3}}}~dx}&=&\displaystyle {{\frac {1}{3}}\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{9}}u^{\frac {3}{2}}+C}\\&&\\&=&\displaystyle {{\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C.}\end{array}}$

Final Answer:
${\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C$

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