Calc Quiz of Death

 $\frac{dy}{dx}=\frac{x^2y+y^2}{4x+2y}$  will have vertical slope field segments when...

•  $y=2x$  

•  $y=-2x$  

•  $y=x^2$  

•  $y=-x^2$  

The equation for this slope field is...

• $\frac{dy}{dx}=xy+y$

• $\frac{dy}{dx}=\frac{x-1}{y}$

• $\frac{dy}{dx}=-x+y$

• $\frac{dy}{dx}=xy-y$

This slope field represents...

• $\frac{dy}{dx}=e^{x+y}$

• $\frac{dy}{dx}=\left(x+y\right)^e$

• $\frac{dy}{dx}=\ln\left(x+y\right)$

• $\frac{dy}{dx}=ex+ey$

What are the requirements for the extreme value theorem?

• Continuous and differentiable on the closed interval

• Continuous on the closed interval and differentiable on the open interval

• Continuous on the open interval, and differentiable on the closed interval

• Continuous and differentiable on the open interval

 $\int_0^x\tan tdt$  

•  $\sin x+C$  

•  $-\ln\cos x+C$  

•  $-\cos x\ +C$  
  

•  $-\ln\left|\cos x\right|+C$  

 $\int_0^x\frac{1+\cos t}{\sin^2t}dt$  

•  $\frac{\cos x-1}{\sin x}+C$  

•  $\frac{-\sin x-\cos x}{\sin^2x}+C$  

•  $-\frac{\left(1+\cos x\right)}{\sin x}+C$  

•  $\frac{\left(\cos x-\sin x\right)}{\sin^2x}+C$  

 $\int_0^x\frac{1}{\sqrt{25-x^2}}dx$  

•  $\sin^{-1}\left(\frac{x}{5}\right)+C$  

•  $\tan^{-1}\left(\frac{x}{5}\right)+C$  

•  $\frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right)+C$  

•  $\frac{1}{5}\sin^{-1}\left(\frac{x}{5}\right)+C$  

Find the first quadrant area bounded by the y-axis,  $y=e^{\frac{x}{2}}$  and the line x=2

•  $2e$  

•  $\frac{e}{2}-1$  

•  $\frac{e-1}{2}$  

•  $2e-2$  

Using the substitution  $u=\sqrt{x}$  ,  $\int_1^4\frac{e^{\sqrt{x}}}{\sqrt{x}}dx$  is equal to which of the following?

•  $2\int_1^{16}e^udu$  

•  $2\int_1^2e^udu$  

•  $\frac{1}{2}\int_1^2e^udu$  

•  $\int_1^4e^udu$  

If $f=\int_6^{12}f\left(2x\right)dx$  , what must be true?

•  $\int_{12}^{24}f\left(t\right)dt=5$  

•  $\int_{12}^{24}f\left(t\right)dt=20$  

•  $\int_3^6f\left(t\right)dt=5$  

•  $\int_6^{12}f\left(t\right)dt=20$