test

 

 

 

\(\boldsymbol {U}\)-Sub Integral Setup Solution and Check
\(\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx\) \(\displaystyle \begin{array}{c}\color{blue}{{u=2{{x}^{2}}+3}}\\du=4x\,dx\end{array}\)

\(\displaystyle dx=\frac{{du}}{{4x}}\)

\(\require {cancel} \displaystyle \int{{{{{\left( {\color{blue}{{2{{x}^{2}}+3}}} \right)}}^{5}}}}x\,\color{green}{{dx}}=\int{{{{{\color{blue}{u}}}^{5}}}}\cancel{x}\color{green}{{\frac{{du}}{{4\cancel{x}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{5}}\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{6}{{\color{blue}{u}}^{6}}+C=\frac{1}{{24}}{{\color{blue}{u}}^{6}}+C\)

\(\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C\)

 

Differentiate back to verify:

\(\displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}} \right)&=6\cdot \frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot 4x\\&=\frac{1}{{\cancel{4}}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot \cancel{4}x={{\left( {2{{x}^{2}}+3} \right)}^{5}}x\end{align}\)    √

\(\int{{\cos \left( {4x} \right)\,{{{\sin }}^{3}}}}\left( {4x} \right)\,dx\)

 

 

Note that this is actually:

\(\displaystyle \int{{\cos \left( {4x} \right)\,}}{{\left( {\sin \left( {4x} \right)} \right)}^{3}}dx\)

\(\displaystyle \begin{align}&\color{blue}{{u=\sin \left( {4x} \right)}}\,\,\\du&=4\cos \left( {4x} \right)\,dx\\&\color{green}{{dx=\frac{{du}}{{4\cos \left( {4x} \right)}}}}\end{align}\)  

\(\displaystyle \begin{align}\int{{\cos \left( {4x} \right){{{\color{blue}{{\sin }}}}^{3}}\left( {\color{blue}{{4x}}} \right)}}\,\color{green}{{dx}}&=\int{{\cancel{{\cos \left( {4x} \right)}}{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{\frac{{du}}{{4\cancel{{\cos \left( {4x} \right)}}}}}}\\&=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{3}}\,\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{4}{{\color{blue}{u}}^{4}}+C=\frac{1}{{16}}{{\color{blue}{u}}^{4}}+C\end{align}\)

\(\displaystyle \int{{\cos \left( {4x} \right){{{\sin }}^{3}}\left( {4x} \right)}}\,dx=\frac{1}{{16}}{{\sin }^{4}}\left( {4x} \right)+C\)

 

Differentiate back to verify:

\(\displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{1}{{16}}{{{\sin }}^{4}}\left( {4x} \right)} \right)&=\frac{1}{{16}}\cdot 4\sin {{\left( {4x} \right)}^{3}}\cdot \cos \left( {4x} \right)\cdot 4\\&=\cancel{{\frac{{16}}{{16}}}}\sin {{\left( {4x} \right)}^{3}}\cos \left( {4x} \right)=\cos \left( {4x} \right)\sin {{\left( {4x} \right)}^{3}}\end{align}\)       √

\(\displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}\) \(\begin{align}\color{blue}{{u=3{{x}^{2}}+2}}\\du=6x\,dx\\\color{green}{{dx=\frac{{du}}{{6x}}}}\end{align}\)

\(\displaystyle \begin{align} \int{{\frac{{x\,\color{green}{dx}}}{{\sqrt{{\color{blue}{{3{{x}^{2}}+2}}}}}}}}&=\int{{{{{\left( {\color{blue}{{3{{x}^{2}}+2}}} \right)}}^{{-\frac{1}{2}}}}x\,\color{green}{{dx}}=}}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}}}\cancel{x}\color{green}{{\frac{{du}}{{6\cancel{x}}}}}=\frac{1}{6}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}\color{green}{du}}}\\&=\frac{{\left( {\frac{1}{6}} \right)}}{{\left( {\frac{1}{2}} \right)}}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C=-\frac{1}{3}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C\end{align}\)

\(\displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}x\,dx=\frac{1}{3}{{\left( {3{{x}^{2}}+2} \right)}^{{\frac{1}{2}}}}+C=\frac{{\sqrt{{3{{x}^{2}}+2}}}}{3}\,+\,C\)

 

Differentiate back to verify:

\(\displaystyle \frac{{d\left( {\frac{{\sqrt{{3{{x}^{2}}+2}}}}{3}} \right)}}{{dx}}\,=\frac{{d\left( {\frac{1}{3}{{{\left( {3{{x}^{2}}+2} \right)}}^{{\frac{1}{2}}}}} \right)}}{{dx}}=\frac{1}{{\cancel{6}}}{{\left( {3{{x}^{2}}+2} \right)}^{{-\frac{1}{2}}}}\cdot \cancel{6}x=\frac{x}{{\sqrt{{3{{x}^{2}}+2}}}}\,\)      √

 

 

\(\boldsymbol {U}\)-Sub Integral

 

\(u=g\left( x \right)\) \(\displaystyle du={g}’\left( x \right)dx\) Substitute, Integrate, and Substitute Back!

\(\displaystyle \int{{{{{\left( {\color{blue}{{5{{x}^{2}}+3}}} \right)}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}\)

\(u=5{{x}^{2}}+3\) \(\displaystyle du=\left( {10x} \right)dx\) \(\displaystyle \int{{{{{\color{blue}{{\left( {5{{x}^{2}}+3} \right)}}}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}=\int{{{{{\color{blue}{u}}}^{4}}}}\,\color{red}{{du}}=\frac{1}{5}{{u}^{5}}+C=\frac{1}{5}{{\left( {5{{x}^{2}}+3} \right)}^{5}}+C\)

\(\displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}\)

\(u=\tan x\) \(\displaystyle \int{{{{{\left( {5{{x}^{2}}+3} \right)}}^{4}}}}\left( {10x} \right)dx\) \(\displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}=\int{{\sqrt{{\color{blue}{u}}}\,}}\color{red}{{du}}=\int{{{{{\color{blue}{u}}}^{{\frac{1}{2}}}}}}\,\color{red}{{du}}=\frac{2}{3}{{u}^{{\frac{3}{2}}}}+C=\frac{2}{3}{{\left( {\tan x} \right)}^{{\frac{3}{2}}}}+C\)

\(\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C\)

\color{red}

\(\require{cancel}  \displaystyle \color{red}{\cancel{{du=\left( {10x} \right)\,dx}}}\)

 

Find the center and radius of the following circle:   \({{x}^{2}}+{{y}^{2}}-6x-12y-55=0\).

Solution:

You may see a “trick” problem like this, where the solution is just a point:

<
<

Problem:

Complete the square and graph:   \({{x}^{2}}+{{y}^{2}}-4x+2y+5=0\).

Solution:

Completing the Square for a Circle Solution

 \(\begin{align}\color{#800000}{{{{x}^{2}}+{{y}^{2}}-4x+2y+5=0}}\\{{x}^{2}}-4x+{{y}^{2}}+2y=-5\end{align}\)

\(\begin{align}\left( {{{x}^{2}}-4x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)+\left( {{{y}^{2}}+2y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=-5+\,\,\underline{{\,\,\,\,\,\,}}\\\left( {{{x}^{2}}-4x+\underline{{{{{\left( 2 \right)}}^{2}}}}\,} \right)+\left( {{{y}^{2}}+2y+\underline{{{{{\left( 1 \right)}}^{2}}}}\,} \right)&=-5+\underline{{{{{\left( 2 \right)}}^{2}}+{{{\left( 1 \right)}}^{2}}}}\\{{\left( {x-\underline{2}} \right)}^{2}}+{{\left( {y+\underline{1}} \right)}^{2}}&=-5+4+1\\{{\left( {x-2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}&=0\end{align}\)

We move the constant to the right side, and we are ready to complete the square!

 

Now we need to divide the coefficients (both \(x\) and \(y\)) of the middle terms by 2 and square them to complete the square. We add the squared constants to the other side.

 

Now we have the equation in circle form, but our radius is 0!

 

So, for this equation, the only solution is a point at \((2,-1)\) (where the center of the circle would normally be). You would just graph this point!

  \(x<-1\) \(x=-1\) \(-1<x<2\) \(x=2\) \(2<x<5\) \(x=5\) \(5<x<8\) \(x=8\)
\(f\left( x \right)\) negative undefined positive 1 –2 1
\({f}’\left( x \right)\) negative undefined negative 0 negative undefined positive
\({{f}^{\prime \prime}(x)}\) negative undefined positive 0 negative negative

 

 

Problem Solution

 

<

 

 

 

 

<table

Vector ProblemSolution

Find x so that the following vectors are orthogonal:

(a)

(b)

We’ll need to find  and make sure it equals 0:

(a)

(b)

Vector Parametric Problem Solution
Find the equation of the plane that passes through the point \(\left( {4,0,-1} \right)\) and is perpendicular to the vector \(n=\text{i}-3\text{j}+2\text{k}\).  

 

The equation of this plane is \(\displaystyle x-3y+2z=2\).

Find the equation of the plane that passes through the point \(\left( {3,-3,1} \right)\), and is perpendicular to the line \(\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}\). We know the direction vector of this line is \(\left\langle {-3} \right.,\left. {4,2} \right\rangle \) (we have to set each expression to \(t\), and solve back for\(x\), \(y\), and \(z\). Also note that we had to multiply the \(x\) expression by –1).

 

 

 

 

The equation of this plane is \(\displaystyle -3x+4y+2z=-19\).

 

Quadratic with Complex Solutions Solve Using Quadratic Formula

\({{x}^{2}}-2x+2=0\)

 

\(\begin{array}{l}a=1\\b=-2\\c=2\end{array}\)

\(\require{cancel} \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 1 \right)\left( 2 \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\pm \sqrt{{-4}}}}{2}=\frac{{2\pm 2i}}{2}\\&=\frac{{\cancel{2}(1\pm i)}}{{\cancel{2}}}=1\pm i\end{align}\)

\(3{{x}^{2}}-2x+2=0\)

 

\(\begin{array}{l}a=3\\b=-2\\c=2\end{array}\)

\(\displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 3 \right)\left( 2 \right)}}}}{{2\left( 3 \right)}}\\&=\frac{{2\pm \sqrt{{-20}}}}{6}=\frac{{2\pm 2\sqrt{5}\,i}}{6}\\&=\frac{{{}^{1}\cancel{2}(1\pm \sqrt{5}\,i)}}{{{{{\cancel{6}}}^{3}}}}=\frac{1}{3}\pm \frac{{\sqrt{5}i}}{3}\text{ }\text{ or }\text{ }\frac{1}{3}\pm \frac{{\sqrt{5}}}{3}i\text{ }\end{align}\)

 

 

<