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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)}   is the lower function.

Solution:

(a)

Step 1:
We use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=1+x^3.}
Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=3x^2dx}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{3}=x^2dx.}
Therefore, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{3}\int \sqrt{u}~du.}

Step 2:
We now have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

(b)

Step 1:
We use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\sin(x).}
Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=\cos(x)dx.}
Also, we need to change the bounds of integration.
Plugging in our values into the equation  $u=\sin(x),$  we get
$u_{1}=\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}$  and  $u_{2}=\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}=1.$
Therefore, the integral becomes
$\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du.$

Step 2:
We now have
${\begin{array}{rcl}\displaystyle {\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx}&=&\displaystyle {\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {\left.{\frac {-1}{u}}\right|_{\frac {\sqrt {2}}{2}}^{1}}\\&&\\&=&\displaystyle {-{\frac {1}{1}}-{\frac {-1}{\frac {\sqrt {2}}{2}}}}\\&&\\&=&\displaystyle {-1+{\sqrt {2}}.}\end{array}}$

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

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