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:

Step1: Calculate \(A(0)\)

By the definition of the definite integral, if \(A(x)=\int_{0}^{x}v(t)dt\), then when \(x = 0\), \(A(0)=\int_{0}^{0}v(t)dt\).
According to the property of definite integrals \(\int_{a}^{a}f(x)dx = 0\), so \(A(0)=0\).

Step2: Calculate \(A(-1)\)

\(A(-1)=\int_{0}^{-1}v(t)dt=-\int_{-1}^{0}v(t)dt\).
The region from \(t=-1\) to \(t = 0\) is a triangle with base \(b = 1\) and height \(h=- 3\). The area of a triangle is \(A=\frac{1}{2}\times base\times height\). So \(\int_{-1}^{0}v(t)dt=\frac{1}{2}\times1\times(-3)=-\frac{3}{2}\), and \(A(-1)=\frac{3}{2}\).

Step3: Calculate \(A(2)\)

\(A(2)=\int_{0}^{2}v(t)dt\).
The region from \(t = 0\) to \(t=1\) is a triangle with base \(b = 1\) and height \(h = 3\) (since the function goes from \(v(0)=-3\) to \(v(1) = 0\)), and the area is \(\frac{1}{2}\times1\times3=\frac{3}{2}\). The region from \(t = 1\) to \(t=2\) is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a = 0\), \(b = 4\), \(h = 1\). The area is \(\frac{(0 + 4)\times1}{2}=2\). So \(A(2)=\frac{3}{2}+2=\frac{3 + 4}{2}=\frac{7}{2}\).

Step4: Calculate \(A(4)\)

\(A(4)=\int_{0}^{4}v(t)dt\).
\(A(2)=\frac{7}{2}\). The region from \(t = 2\) to \(t=4\) is a trapezoid. Using the trapezoid - area formula \(A=\frac{(a + b)h}{2}\), where \(a = 4\), \(b = 0\), \(h = 2\). The area is \(\frac{(4+0)\times2}{2}=4\). So \(A(4)=\frac{7}{2}+4=\frac{7 + 8}{2}=\frac{15}{2}\).

Step5: Find \(A^{\prime}(2)\)

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, \(v(2) = 4\).

Step6: Determine when \(A(x)\) is increasing or decreasing

\(A(x)\) 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 \(x\in(-\infty,1)\cup(4,\infty)\).

Step7: Find the position \(s(4)\)

We know that \(s(t)=s(0)+\int_{0}^{t}v(u)du\). Given \(s(0) = 5\) and \(A(4)=\int_{0}^{4}v(t)dt=\frac{15}{2}\). Then \(s(4)=s(0)+A(4)\). Substitute \(s(0) = 5\) and \(A(4)=\frac{15}{2}\), we get \(s(4)=5+\frac{15}{2}=\frac{10 + 15}{2}=\frac{25}{2}\)

Answer:

• \(A(0)=0\), \(A(-1)=\frac{3}{2}\), \(A(2)=\frac{7}{2}\), \(A(4)=\frac{15}{2}\)
• \(A^{\prime}(2)=4\)
• \(A(x)\) is increasing on \((1,4)\) and decreasing on \((-\infty,1)\cup(4,\infty)\)
• \(s(4)=\frac{25}{2}\) feet