AP Calculus BC Final

Find the indefinite integral.
 $\int x^3\ \ln\left(x\right)\ dx$  

•  $\frac{x^4}{9}\ \left[\ln\left(x^4\right)-1\right]+C$  

•  $\frac{x^3}{16}\ \left[\ln\left(x^3\right)-1\right]+C$  

•  $\frac{x^2}{16}\ \left[\ln\left(x^2\right)-1\right]+C$  

•  $\frac{x^4}{16}\ \left[\ln\left(x^4\right)-1\right]+C$  

•  $\frac{x^4}{16}\ \left[\ln\left(x^3\right)-1\right]+C$  

Use partial fractions to find the integral.
 $\int\ \frac{13x-109}{x^2-13x+30}dx$  

•  $10\ \ln\left|x-3\right|+3\ln\ \left|x-10\right|+C$  

•  $10\ x+30\ln\left|3-x\right|+3x+30\ln\ \left|10-x\right|+C$  

•  $10\ \ln\left|x-3\right|-3\ln\ \left|x-10\right|+C$  

•  $10\ x+30\ln\left|3-x\right|-3x+30\ln\ \left|10-x\right|+C$  

•  $10x+30\ \ln\left|3-x\right|+3\ln\ \left|10-x\right|+C$  

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges.
 $\int_0^1\ \frac{1}{\left(1-x\right)^{\frac{1}{9}}}\ dx$  

• diverges

•  $converges\ to\ \frac{9}{8}$  

•  $converges\ to\ 9$  

•  $converges\ to\ \frac{8}{9}$  

•  $converges\ to\ 8$  

Find the sum of the convergent series.
 $\sum_{n=1}^{\infty}\frac{3}{\left(n+9\right)\left(n+11\right)}$  

•  $\frac{117}{220}$  

•  $\frac{19}{60}$  

•  $\frac{45}{143}$  

•  $\frac{63}{220}$  

•  $\frac{59}{110}$  

Find the Maclaurin polynomial of degree 4 for the function.

 $f\left(x\right)=\cos\left(3x\right)$  

•  $1+\frac{9}{2}x^2-\frac{27}{8}x^4$  

•  $1-\frac{9}{2}x^2+\frac{81}{40}x^4$  

•  $1-\frac{9}{2}x^2+\frac{27}{8}x^4$  

•  $1+\frac{9}{2}x^2-\frac{81}{40}x^4$  

•  $x-\frac{9}{2}x^3-\frac{81}{40}x^5$  

Find a geometric power series for the function centered at 0.
 $f\left(x\right)=\frac{5}{9-x}$  

•  $\sum_{n=0}^{\infty}5\left(\frac{-x}{9}\right)^n\ \ ,\ \ \ \left|x\right|<9$  

•  $\sum_{n=0}^{\infty}\frac{5}{9}\left(-x\right)^n\ \ ,\ \ \ \left|x\right|<1$  

•  $\sum_{n=0}^{\infty}\frac{5}{9}\left(-9x\right)^n\ \ ,\ \ \ \left|x\right|<9$  

•  $\sum_{n=0}^{\infty}\frac{5}{9}\left(\frac{x}{9}\right)^n\ \ ,\ \ \ \left|x\right|<9$  

• None of the above

Use the definition to find the Taylor series (centered at c) for the function.
 $f\left(x\right)=\cos\left(x\right)\ ,\ c=\frac{\pi}{4}$  

•  $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2\left(2!\right)}\left(x-\frac{\pi}{4}\right)^2-\frac{\sqrt{2}}{2\left(3!\right)}\left(x-\frac{\pi}{4}\right)^3-.....$  

•  $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2\left(2!\right)}\left(x-\frac{\pi}{4}\right)^2+\frac{\sqrt{2}}{2\left(3!\right)}\left(x-\frac{\pi}{4}\right)^3-.....$  

•  $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)+\frac{\sqrt{2}}{2\left(2!\right)}\left(x-\frac{\pi}{4}\right)^2-\frac{\sqrt{2}}{2\left(3!\right)}\left(x-\frac{\pi}{4}\right)^3+.....$  

•  $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2\left(2!\right)}\left(x-\frac{\pi}{4}\right)^2+\frac{\sqrt{2}}{2\left(3!\right)}\left(x-\frac{\pi}{4}\right)^3+.....$  

•  $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2\left(2!\right)}\left(x-\frac{\pi}{4}\right)^2-\frac{\sqrt{2}}{2\left(3!\right)}\left(x-\frac{\pi}{4}\right)^3-.....$  

Convert the polar equation to rectangular form.

r=5

• $x=5$

• $x^2+\left(y-5\right)^2=25$

• $x^2+y^2=25$

• $6x-y+5=0$

• $y=5$

 $Find\ \frac{\text{d}y}{\text{d}x}and\ \frac{\text{d}^2y}{\text{d}x^2}and\ find\ the\ slope.$  

 $Find\ the\ concavity\ corresponding\ to\ t=3.$  

 $x=t+7\ \ \ y=t^2+12t$  

•  $\frac{\text{d}y}{\text{d}x}=2t+12,\ \frac{\text{d}^2y}{\text{d}x^2}=2.\ At\ t=3:\ slope\ 18\ and\ concave\ up$  

•  $\frac{\text{d}y}{\text{d}x}=-2t+12,\ \frac{\text{d}^2y}{\text{d}x^2}=-2.\ At\ t=3:\ slope\ 6\ and\ concave\ down$  

•  $\frac{\text{d}y}{\text{d}x}\ \frac{1}{3}t^3+6t^2,\ \frac{\text{d}^2y}{\text{d}x^2}=3t^2+12t.\ At\ t=3:\ slope\ \frac{27}{2}\ and\ concave\ up$  

•  $\frac{\text{d}y}{\text{d}x}=-2t+12,\ \frac{\text{d}^2y}{\text{d}x^2}=-2.\ At\ t=3:\ slope\ -6\ and\ concave\ down$  

•  $\frac{\text{d}y}{\text{d}x}=\frac{1}{3}t^3+6t^2,\ \frac{\text{d}^2y}{\text{d}x^2}=3t^2+12t.\ At\ t=3:\ slope\ -\frac{27}{2}\ and\ concave\ up$  

 $Find\ \parallel r\left(t\right)\parallel\ given\ the\ function\ r\left(t\right)\ below.$  
 $r\left(t\right)=2\sin5t\ i\ +2\cos5t\ j\ +11t\ k$  

•  $\sqrt{4+11t}$  

•  $4+121t^2$  

•  $2+11t$  

•  $\sqrt{2+121t^2}$  

•  $\sqrt{4+121t^2}$