${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(4x^3+cosx)dx=$

A$.x^4+sinx+C$

B$.x^4-sinx+C$

C$.12x^2+sinx+C$

D$.12x^2-sinx+C$

What is ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(\frac{1}{x}-se{c}^{2}x)dx?$

A$.-\frac{1}{x^2}-tanx+C$

B$.-ln|x|-2se{c}^{2}xtanx+C$

C$.lnx+tanx+C$

D$.lnx-tanx+C$

The positive value of $z$ for which ${\displaystyle \underset{z}{\overset{5}{}}}2xdx=9$ is

A$.3$

B$.4$

C$.6$

D$.8$

What is ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x^3}dx?$

A$.\frac{3}{2}\sqrt[\phantom{2}]{x^5}+C$

B$.\frac{2}{5}\sqrt[\phantom{2}]{x^5}+C$

C$.\frac{2}{5}{x}^{\frac{1}{5}}+C$

D$.\frac{3}{2}\sqrt[\phantom{2}]{x}+C$

The gradient of the normal to the curve to the curve $y=x^2-3x+2$ at the point where $x=0$ is :

A$.-3$

B$.0$

C$.\frac{1}{3}$

D$.-\frac{1}{3}$

The area of the region bounded by the curve $y=x^2$ and the lines $x=1$ and $x=2$ and the $x$ axis is:

A$.\frac{3}{7}$ square units

B$.\frac{8}{3}$ square units

C$.\frac{7}{3}$ square units

D$.3$ square units

${\displaystyle \underset{17}{\overset{21}{}}}xdx$

A$.19$

B$.76$

C$.38$

D$.152$

Let $R$ be the triangular region bounded by the $x-$axis, the $y-$axis, and the line $y=1-\frac{1}{2}x.$ When $R$ is revolved around the $x-$axis, you will get the same volume as when is revolved around the $y-$axis.

A$.$ True

B$.$ False