Quiz 2 (Solution)

Date: MATH 1210-4 - Spring 2004

1. Find $\int (x^3+3x^2-1) dx$.

   $$\begin{split}\int (x^3+3x^2-1) dx &= \int x^3 dx + \int 3x^2 ... ...rac{1}{3} x^3 - x +C = \frac{1}{4}x^4 + x^3 - x +C. \end{split}$$

   
   

   We can check the result: $(\frac{1}{4}x^4 + x^3 - x +C)' = (\frac{1}{4}x^4)' + (x^3)' - (x)' +(C)' = \frac{1}{4}\cdot 4 x^3 + 3x^2-1+0 = x^3+ 3x^2-1$, as wanted.

2. Find the area determined by the $x$-axis and the curve $y=6x^5-4x$ for $2\leq x\leq 5$.

   Since $6x^5-4x\geq 0$ for $2\leq x\leq 5$ we can compute the area as $\int_2^5( 6x^5-4x) dx$:

   $$\begin{split}Area &= \int_2^5( 6x^5-4x) dx = (x^6 - 2x^2)\ver... ...2^6 - 2\cdot 2^2) \\ &= 15625 -50 -(64 -8) = 15519. \end{split}$$

   
   

3. A ball is thrown straight up from the top of a $100 ft$ tall building with an initial velocity of $16 ft/s$. The acceleration due to the gravity force is $-32 ft/s^2$.
   1. Find a formula for the velocity of the ball $t$ seconds after it is thrown.

      Since $a=v'$ we have $v(t) = \int a(t) dt = \int -32 dt = -32 t+ C$. But we know that $v(0) = 16$, so that $v(0) = 0+C = 16 \ensuremath{\Rightarrow}\xspace C=16$ and we have $v(t) = -32t+16$.

   2. Find a formula for the position (height above the ground) of the ball $t$ seconds after it is thrown.

      We denote the height above the ground after $t$ seconds with $s(t)$. Then, since $v=s'$ we have $s(t) = \int v(t) dt = \int (-32 t+16) dt = -32 \frac{t^2}{2} + 16 t + C$. Additionally, $s(0) = 100$, so that $100 = s(0) = C$ and the final formula is $s(t) = -16t^2 + 16 t + 100$.

   3. What is the maximum height that the ball reaches?

      The time $t_{max}$ of the maximum height satisfies $v(t_{max}) = 0$. That is, $0 = -32t_{max}+16$, so that $t_{max}=\frac{1}{2}$.

      The position at $t_{max}$, the maximum height, is $s(\frac{1}{2}) = -16(\frac{1}{2})^2 + 16 \cdot \frac{1}{2} + 100 = -4 + 8 + 100 = 104$ ft above the ground.