13. let ( a(x)=int_{0}^{x} v(t) d t ).\na. find the following values:\n( a(0)= )( a(-1)= )( a(2)= )( a(4)=…

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.