Question

1. let ( a(x)=int_{0}^{x} v(t) d t ).

a. find the following values:
( a(0)= )( a(-1)= )( a(2)= )( a(4)= )
b. find ( a^{prime}(2) ).
c. when is ( a(x) ) increasing, and when is ( a(x) ) decreasing?
d. suppose that ( s(t) ) is the position in feet of a particle at time ( t ) seconds. suppose that ( s^{prime}(t)=v(t) ) is the velocity of the particle in feet/second at time ( t ) seconds. if ( s(0)=5 ) feet, find the position of the particle at time 4 seconds.

Explanation:

Part A:

For \(A(0)\):

By the definition of the definite integral \(\int_{a}^{a}f(x)dx = 0\). Here, \(A(x)=\int_{0}^{x}v(t)dt\), so when \(x = 0\), \(A(0)=\int_{0}^{0}v(t)dt=0\).

For \(A(- 1)\):

The integral \(\int_{0}^{-1}v(t)dt=-\int_{-1}^{0}v(t)dt\). The region from \(t=-1\) to \(t = 0\) is a triangle with base \(b = 1\) and height \(h=- 4\). The area of a triangle is \(A=\frac{1}{2}\times base\times height\). So \(\int_{-1}^{0}v(t)dt=\frac{1}{2}\times1\times(-4)=-2\). Then \(A(-1)=-\int_{-1}^{0}v(t)dt = 2\).

For \(A(2)\):

\(A(2)=\int_{0}^{2}v(t)dt=\int_{0}^{1}v(t)dt+\int_{1}^{2}v(t)dt\).

• The region from \(t = 0\) to \(t = 1\) is a triangle with base \(b = 1\) and height \(h = 4\). The area \(\int_{0}^{1}v(t)dt=\frac{1}{2}\times1\times4 = 2\).
• The region from \(t = 1\) to \(t = 2\) is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{1}{2}(a + b)h\), where \(a = 4\), \(b\) (the value of \(v(2)\)) can be found from the slope. The slope from \(t = 1\) to \(t = 2\): The line from \(t=1\) (\(v(1) = 0\)) to \(t = 2\) (let \(v(2)=y\)). The slope of the line \(v(t)\) from \(t = 1\) to \(t=2\) is \(m = 4\) (since from \((1,0)\) to \((2,4)\)), so \(v(2)=4\). The area \(\int_{1}^{2}v(t)dt=\frac{1}{2}(0 + 4)\times1=2\).

So \(A(2)=2 + 2=4\).

For \(A(4)\):

\(A(4)=\int_{0}^{4}v(t)dt=\int_{0}^{2}v(t)dt+\int_{2}^{4}v(t)dt\).

• We know \(\int_{0}^{2}v(t)dt = 4\).
• The region from \(t = 2\) to \(t = 4\) is a trapezoid. \(v(2)=4\), \(v(4)=0\), and \(h = 2\). Using the trapezoid area formula \(A=\frac{1}{2}(a + b)h\), \(\int_{2}^{4}v(t)dt=\frac{1}{2}(4+0)\times2=4\).

So \(A(4)=4 + 4=8\).

Part B:

By the Fundamental Theorem of Calculus, if \(A(x)=\int_{0}^{x}v(t)dt\), then \(A^{\prime}(x)=v(x)\). So \(A^{\prime}(2)=v(2)\). From the graph (or by calculating the slope of the line segment from \(t = 1\) to \(t = 2\): using two - point form \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(1,0)\) and \((x_2,y_2)=(2,4)\), \(m = 4\)), \(v(2)=4\).

Part C:

By the Fundamental Theorem of Calculus \(A^{\prime}(x)=v(x)\).

• \(A(x)\) is increasing when \(A^{\prime}(x)=v(x)>0\). From the graph, \(v(x)>0\) for \(x\in(1,4)\).
• \(A(x)\) is decreasing when \(A^{\prime}(x)=v(x)<0\). From the graph, \(v(x)<0\) for \(x\in(-\infty,1)\cup(4,\infty)\).

Part D:

We know that \(s(t)=s(0)+\int_{0}^{t}v(u)du\). Given \(s(0) = 5\) and \(t = 4\). Since \(A(4)=\int_{0}^{4}v(t)dt = 8\). Then \(s(4)=s(0)+\int_{0}^{4}v(t)dt\). Substituting the values, \(s(4)=5 + 8=13\) feet.

Answer:

• \(A(0)=0\), \(A(-1)=2\), \(A(2)=4\), \(A(4)=8\)
• \(A^{\prime}(2)=4\)
• \(A(x)\) is increasing on \((1,4)\) and decreasing on \((-\infty,1)\cup(4,\infty)\)
• The position of the particle at \(t = 4\) seconds is \(13\) feet.