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5.3 The Fundamental Theorem of Calculus Part 2

5.3.1 Two Proofs of The Fundamental Theorem of Calculus Part 2

 

We provide two proofs of TFTCP2. One using the MVT (for derivatives) and telescoping series; the second one as a direct consequence of TFTCP1, plus an application of a consequence of the MVT.

Theorem 1. [The Fundamental Theorem of Calculus Part 2] Let \(f\) be a continuous function on an open interval \(I\), and assume that \(F\) is an antiderivative of \(f\) on \(I\). Then, for each \(a, b\in I\),
\begin{equation}\label{ch05-03-eq01}\tag{5.4}
\int_a^bf(x)\,dx = F(b)-F(a).
\end{equation}

Proof 1.
By definition of the definite integral, we have
\[\int_a^bf(x)\,dx \overset{\text{Def.}}{=}\lim_{
\renewcommand{\arraystretch}{.7} \begin{array}{c}
{\scriptstyle n\to \infty }\\ {\scriptstyle \Delta x_i\to 0 }
\end{array}
}\sum_{i=1}^nf(x_i^*)\Delta x_i,\]
where \(x_i^*\) is a random point on the \(i\)-th interval. By the Existence of Definite Integral Theorem (Theorem 1 in 3.4.1), to prove TFTCP2 we need to show that for at least one sequence of Riemann sums the limits is precisely \(F(b)-F(a)\).
The proof that follows constructs one such Riemann sum.

Step 1. Begin with a partition of the interval \([a,b]\) into \(n\) subintervals of equal length:
\[\Delta x =\frac{b-a}{n},\, x_i=a+i \Delta x,\, i=0,1,2, \dots n, \text{ for } n \in \mathbb{N} \text{ and } n>1.\]

Setp 2. Construct a telescoping sum as follows:
\[ \sum_{i=0}^n b_i, \;\; \text{with}\;\; b_i=F(x_i)-F(x_{i-1}).\]
Note that \(b_0=F(x_1)-F(x_0)=F(x_1)-F(a)\), and \(b_n=F(x_n)-F(x_{n-1})=F(b)-F(x_{n-1})\). Then
\begin{align*}
\sum_{i=0}^n b_i &=(F(x_1)-F(x_0)) + (F(x_2)-F(x_1))+ (F(x_3)-F(x_2))+ \cdots +\\
&\hspace{.5in}+(F(x_{n-1})-F(x_{n-2})) + (F(x_n)-F(x_{n-1})) = F(x_n)-F(x_0) = F(b)-F(a).
\end{align*}

Setp 3. Apply the Mean Value Theorem (MVT) to \(F\) on each interval \([x_{i-1},x_i]\).

First of all, note that \(F\) satisfies the hypothesis of the MVT on each subinterval \([x_{i-1},x_i]\subset I\).

Since \(F\) is differentiable on \(I\), thus, in particular, it is also continuous on \(I\). It follows that \(F\) is continuous on \([x_{i-1},x_i]\), and differentiable on \((x_{i-1},x_i)\).

We can now apply the MVT to conclude that for each interval \([x_{i-1},x_i]\), there exists a number \(x_i^* \in (x_{i-1},x_i)\) such that
\[F'(x_i^*)= \frac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}, \;\; \text{that is,} \;\; F(x_i)-F(x_{i-1}) = F'(x_i^*)(x_i-x_{i-1}), \]
but, since \(f'(x)=f(x)\) for all \(x\in I\), as \(F\) is an antiderivative of \(F\), then
\[F(x_i)-F(x_{i-1})= F'(x_i^*)(x_i-x_{i-1}) = f(x_i^*)(x_i-x_{i-1}).\]

Step 4. Substitute into the telescoping sum to obtain a Riemann sum:
\begin{align*}
F(b)-F(a)
&\;\; \underset{\text{Telesc. sum}}{=} \;\;\sum_{i=1}^n(F(x_i)-F(x_{i-1}) )
\underset{\text{MVT}}{=} \sum_{i=1}^nF'(x_i^*)(x_i-x_{i-1}) =\\
&
\;\; \underset{\;\;\;\;\; F’=f \;\;\;\;\;}{=} \;\;\sum_{i=1}^nf(x_i^*)(x_i-x_{i-1})
= \sum_{i=1}^nf(x_i^*)\Delta x.
\end{align*}
In the last equality, we have used \(\Delta x = x_i-x_{i-1} = \) length of the intervals of the partition.

Step 5. Taking the limit as \(n\to \infty\),
\[\int_a^bf(x)\,dx = \lim_{n\to \infty} \sum_{i=1}^nf(x_i^*)\Delta x= F(b)-F(a).\]

\(\Box\)

Proof 2.
Let \(\displaystyle G(x) = \int_a^x f(t)\,dt \), then, by TFTCP1, \(G\) is an antiderivative of \(f\). Since \(F\) is assumed to be an anitderivative of \(f\), there exists a constant \(C\) (see Theorem 1) such that
\[F(x) = G(x) + C \; \; \forall \,x \in I. \;\; \Longrightarrow F(b) – F(a) = (G(b) + C) – (G(a) + C ) = G(b)-G(a).
\]
Since \(G(a)= 0\), it follows that
\[F(b) – F(a) = G(b) – G(a) = G(b) = \int_a^bf(x)\,dx .\]

\(\Box\)

Notation. It is standard notation to write
\[\int_a^bf(x)\,dx = F(x) \Big|_a^b = F(b)-F(a).\]

Basic Applications of TFTCP2

 

Example 1. Evaluate \(\displaystyle \int_a^b \sqrt[3]{x^2}\,dx\), \(a, b \in \mathbb{R}\).

Solution. Since \(f(x)= \sqrt[3]{x^2}\) is continuous everywhere, and since \(\displaystyle F(x)= \frac{3}{5}x^{5/3}\) is an antiderivative of \(f\) on \(\mathbb{R}\), it follows that we can apply TFTCP2 to obtain
\[\int_a^b \sqrt[3]{x^2}\,dx = \frac{3}{5}x^{5/3}\Big|_a^b = \frac{3}{5}b^{5/3}-\frac{3}{5}a^{5/3}= \frac{3}{5}b \sqrt[3]{b^2}- \frac{3}{5}a \sqrt[3]{a^2}.\]

Example 2. Evaluate \(\displaystyle \int_{-\pi/6}^{3\pi/4}\sin 2x\,dx\).

Solution. Since \(f(x)=\sin 2x\) is continuous everywhere, and since \(\displaystyle F(x)= -\frac{1}{2}\cos 2x\) is an antiderivative of \(f\) on \(\mathbb{R}\), it follows that we can apply TFTCP2 to obtain
\[\int_{-\pi/6}^{3\pi/4}\sin 2x\,dx= -\frac{1}{2}\cos 2x\Big|_{-\pi/6}^{3\pi/4}
=-\frac{1}{2}\cos\left ( 2\cdot \frac{3\pi}{4} \right ) -\left ( – \frac{1}{2}\cos \left (-2\cdot \frac{\pi}{3}\right )\right )=0 -\left (-\frac{1}{4}\right )=\frac{1}{4}.\]

Example 3. What is wrong with the following \(\displaystyle \int_{0}^{1}\sec^2 \pi x\,dx = \frac{1}{\pi}\tan (\pi\cdot 1) -\frac{1}{\pi}\tan (\pi \cdot 0) =0 \).

Solution. When applying TFTCP2 to evaluate a definite integral \(\displaystyle \int_a^bf(x)\,dx\), we need to verify that all the assumptions of the theorem are satisfied. Namely,

  1. \(f\) is continuous on an open interval \(I\) with \([a,b]\subset I\), and
  2. \(F\) is an antiderivative of \(f\) on \(I\).

Here, (ii) is satisfied, as \(\displaystyle F(x) = \frac{1}{\pi}\tan \pi x\) is an antiderivative of \(f(x)=\sec^2\pi x\). However, \(\sec^2 \pi x\) does not satisfy condition (i), as
\[D_{f}= \left \{x\in \mathbb{R}: x\neq \frac{2n-1}{2}, n \in \mathbb{Z}\right \}.\]
Thus, \(F\) is not continuous at \(\displaystyle x=\frac{1}{2}\), hence, it is not continuous on \([0,1]\). Therefore, the TFTCP2 cannot be applied.

Area and Net Area

 

Example 4. Given the function \(g(x)=2\cos x-1\), \(x\in [0, \pi]\),

(a) sketch the graph of \(g\);

(b) find the net area enclosed by the graph of \(g\) and the \(x\)-axis;

(c) find the area enclosed by the graph of \(g\) and the \(x\)-axis.

Solution. (a) Note that the \(x\)-intercept of the graph is \(\displaystyle \left (\frac{\pi}{3},0\right )\). The graph is shown in Figure 5.1(a).

Shading of the region enclosed by the graph of y = 2cos x – 1 on the interval [0,pi].

Figure 5.1a

Region above x-axis marked +, below -

Figure 5.1b

(b) Solution 1. The net area is given precisely by the definite integral
\[\int_{0 }^{\pi } (2\cos x-1) \,dx = 2\sin x -x \Big|_{0 }^{\pi } = 2\sin \pi-\pi- \left ( 2\sin 0 -0\right ) =-\pi.\]
Solution 2. Alternatively,
since \(f(x)>0\) for all \(x\in\left [0,\frac{\pi}{3}\right )\) (see Figure 5.1(b)), the area above the \(x\)-axis, \(A_+\), is given by
\[
A_+=\int_{0 }^{\pi/3 } (2\cos x-1) \,dx
= 2\sin x -x \Big|_{0 }^{\pi/3 }
= 2\sin \frac{\pi}{3} -\frac{\pi}{3} – \left ( 2\sin 0 -0\right ) =2\frac{\sqrt{3}}{2} -\frac{\pi}{3} =\sqrt{3}-\frac{\pi}{3}.
\]
Since \(f(x)<0\) for all \(x\in\left (\frac{\pi}{3},\pi\right ]\), the area below the \(x\)-axis, \(A_-\), is given by \begin{align*} A_-&=\int_{\pi/3 }^{\pi } - (2\cos x-1) \,dx = -(2\sin x -x) \Big|_{\pi/3 }^{\pi } = \\ & = -\left ((2\sin \pi -\pi) - \left (2\sin \frac{\pi}{3} -\frac{\pi}{3} \right ) \right )=-\left (-\pi -\sqrt{3}+\frac{\pi}{3}\right ) = \sqrt{3}+\frac{2\pi}{3}. \end{align*} Thus, the net area is the difference of areas \[A_+ - A_- =\left (\sqrt{3}-\frac{\pi}{3}\right ) - \left (\sqrt{3}+\frac{2\pi}{3} \right ) = -\pi .\] (c) The area enclosed by the graph of \(g\) and the \(x\)-axis is given (in the notation of part (b)) by \[A = A_+ + A_- = \left (\sqrt{3}-\frac{\pi}{3} \right ) + \left ( \sqrt{3}+\frac{2\pi}{3} \right ) = 2\sqrt{3}+\frac{\pi}{3} .\]

Computing Limits of Riemann Sums via TFTCP2

 

Example 5. Find the exact value of the limit by first identifying it as a Riemann sum, and then applying TFTCP2.
\[\lim_{n\to \infty}\left (\frac{1}{n^2}+\frac{4}{n^2}+\frac{9}{n^2}+ \cdots + \frac{n^2}{n^2}\right )\frac{1}{n}.\]

Solution. The limit can be rewritten using sigma notation as follows:
\[\lim_{n\to \infty}\left (\frac{1}{n^2}+\frac{4}{n^2}+\frac{9}{n^2}+ \cdots + \frac{n^2}{n^2}\right )\frac{1}{n} = \lim_{n\to \infty} \sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n}.\]
We now need to match \(\displaystyle \sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n}\) with the Riemann sum \(\displaystyle \sum_{i=1}^{n}f(x_i)\Delta x\).
\[\Delta x = \frac{b-a}{n} =\frac{1}{n}\Longrightarrow b-a = 1.\]
\[f(x_i)=\frac{i^2}{n^2} =\left (\frac{i}{n}\right )^2\Longrightarrow f(x)=x^2 \;\; \text{and} \;\; x_i=a + i\Delta x = \frac{i}{n}\Longrightarrow a=0 \Longrightarrow b=1.\]
It follows that
\[\lim_{n\to \infty} \sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n} = \int_{0 }^{1 } x^2 \,dx
= \frac{1}{3}x^3 \Big|_{0 }^{1 } = \frac{1}{3}.\]

5.3.2 Exercises

In problems 14, (a) show that \(F\) is an antiderivative of \(f\); (b) use TFTCP2 to evaluate \(\displaystyle \int_a^bf(x)\,dx\) over the interval \(I=[a,b]\).

  1. \(F(x)=e^{\sin^2x}\), \(f(x)=\sin 2 x e^{\sin^2x}\), \(\displaystyle I=\left [0,\frac{\pi}{3}\right ]\).
  2. \(F(x)= \arctan^2 2x \), \(\displaystyle f(x)=\frac{2\arctan 2x}{1+4x^2}\), \(\displaystyle I=\left [0,\frac{\sqrt{3}}{2}\right ]\).
  3. \(\displaystyle F(x)= \ln^2\sqrt[5]{x^4}\), \(\displaystyle f(x)=\frac{32\ln x}{25x}\), \(I=\left [\sqrt{e},e^2\right ]\).
  4. \(\displaystyle F(x)= x^2\arcsin \sqrt{x}\), \(\displaystyle f(x)=\frac{x\sqrt{x}}{2\sqrt{1-x}} + 2x\arcsin \sqrt{x}\), \(\displaystyle I=\left [\frac{1}{4},\frac{3}{4}\right ]\).

In problems 58, use TFTCP2 to evaluate \(\displaystyle \int_a^bf(x)\,dx\) over the interval \(I=[a,b]\).

  1. \(f(x)=\sin \pi x\), \(\displaystyle I=\left [-\frac{1}{2},\frac{3}{4}\right ]\).
  2. \(\displaystyle f(x)=\frac{1+x^2}{x\sqrt{x}}\), \(\displaystyle I=\left [\frac{1}{4},4\right ]\).
  3. \(\displaystyle f(x)=e^{2x}\), \(I=\left [\ln 2,\ln 4\right ]\).
  4. \(\displaystyle f(x)=\frac{2}{\sqrt{1-x^2}}\), \(\displaystyle I=\left [-\frac{1}{2},\frac{\sqrt 3}{2}\right ]\).

In problems 912, explain what is wrong with the evaluation of the definite integral.

  1. \(\displaystyle \int_{-1}^{2}\frac{1}{x^2}\,dx = -\frac{1}{x} \Big |_{-1}^{2}=-\frac{1}{2}-1 = -\frac{3}{2}\).
  2. \(\displaystyle \int_{2}^{4}\frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x \Big |_{2}^{4}= \)
    \(\hphantom{\displaystyle \int_{2}^{4}\frac{1}{\sqrt{1-x^2}}\,dx} = \arcsin 4- \arcsin 2\).
  3. \(\displaystyle \int_{0}^{2}2^x\,dx = 2^x \Big |_{0}^{1}=2-1=1\).
  4. \(\displaystyle \int_{-2}^{-1}\frac{1}{x}\,dx =\ln x\Big |_{-2}^{-1}=\ln(-1)-\ln(-2)\).

In problems 1316, find (a) the net area; (b) the area of the region enclosed by the graph of the function and the \(x\)-axis.

  1. \(f(x)= x^2-2x\), \(x\in[0,3]\).
  2. \(g(x)= 2\sin \pi x-1\), \(x\in [0, 2] \vphantom{\displaystyle f(x)=\frac{1}{x}-1}\).
  3. \(h(x) = e^{2x}-e\), \(x \in [-\ln 2, 2]\).
  4. \(\displaystyle f(x)=\frac{1}{x}-1\), \(\displaystyle x\in \left [\frac{1}{2}, 2\right ]\).

In problems 1722, (a) identify the sum as a Riemann sum and rewrite it using sigma notation. State explicitly \(f(x)\), \([a,b]\), \(\Delta x\), and \(x_i\); (b) write the corresponding definite integral; (c) apply TFTCP2 to find its exact value of the limit.
 

  1. \(\displaystyle \lim_{n\to \infty } \left(\frac{8}{n^3}+\frac{32}{n^3}+\frac{72}{n^3}+\text{…} +\frac{8n^2}{n^3}\right)\).
  2.  

  3. \(\displaystyle \lim_{n\to \infty } \left(\sqrt{\frac{2}{n}}+\sqrt{\frac{4}{n}}+\sqrt{\frac{6}{n}}+\text{…} +\sqrt{\frac{2n}{n}}\right)\frac{2}{n}\).
  4.  

  5. \(\displaystyle \lim_{n\to \infty } \left(
    2\sin\left (\frac{\pi}{6}+ \frac{\pi}{12n}\right )+
    2\sin\left (\frac{\pi}{6}+ \frac{2\pi}{12n}\right ) +
    %2\sin\left (\frac{\pi}{6}+ \frac{3\pi}{12n}\right )+
    \text{…} +
    2\sin\left (\frac{\pi}{6}+ \frac{n\pi}{12n}\right )
    \right)\frac{\pi}{12n}\).
  6.  

  7. \(\displaystyle \lim_{n\to \infty } \left(\sqrt[4]{\frac{1}{n^3}}+\sqrt[4]{\frac{8}{n^3}}+\sqrt[4]{\frac{27}{n^3}}+\text{…} +\sqrt[4]{\frac{n^3}{n^3}}\right)\frac{1}{n}\).
  8.  

  9. \(\displaystyle \lim_{n\to \infty } \left(
    \left (1+ \frac{2}{n}\right )^4+
    \left (1+ \frac{4}{n}\right )^4 +
    \left (1+ \frac{6}{n}\right )^4+\text{…} +
    \left (1+ \frac{2n}{n}\right )^4
    \right)\frac{2}{n}\).
  10.  

  11. \(\displaystyle \lim_{n\to \infty } \left(
    \cot^2\left (\frac{\pi}{6}+ \frac{\pi}{3n}\right )+
    \cot^2\left (\frac{\pi}{6}+ \frac{2\pi}{3n}\right ) +
    \cot^2\left (\frac{\pi}{6}+ \frac{3\pi}{3n}\right )+\text{…} +
    \cot^2\left (\frac{\pi}{6}+ \frac{n\pi}{3n}\right )
    \right)\frac{\pi}{3n}\).
  12.  

  13. Do parts (a) and (b) as in the previous exercises; (c) identify the integral as representing the area of a region, and use a geometric argument to find the value of the limit.
    \[\displaystyle \lim_{n\to \infty } \left(\sqrt{4-\left(-2+\frac{4}{n}\right)^2} + \sqrt{4-\left(-2+\frac{8}{n}\right)^2}+%\sqrt{4-\left(-2+\frac{12}{n}\right)^2}+
    \cdots + \sqrt{4-\left(-2+\frac{4n}{n}\right)^2}\right)\frac{4}{n}.\]
  14. State the definition of a definite integral.
  15. State the Existence Theorem for definite integrals.
  16. State an prove TFTCP2. Show explicitly how telescoping sums, the MVT (for derivatives), and Riemann sums are used in the proof.
  17. State an prove TFTCP1.

5.3.3 The Net Change Theorem

Let \(f\) be a differentiable function defined on an open interval \(I\), and \(x\in I\). Then, we can think of \(f\) as an antiderivative of \(f’\), and hence, TFTCP2 can be interpreted as follows.

Theorem 2. [The Net Change Theorem] Let \(f\) be a differentiable function on an open interval \(I\). Then
\begin{equation}\label{ch05-03-eq02}\tag{5.5}
\int_a^bf'(x)\,dx = f(b)-f(a), \;\;\forall \, a,b\in I.
\end{equation}

Furthermore, the derivative \(f'(x)\) can be interpreted as the rate of change of \(f\) with respect to \(x\).

And, the expression \( f(b) – f(a)\) can be interpreted as the net change of the function \(f\) over the interval \([a,b]\).

Thus, the Net Change Theorem can be stated, in words, as


the integral of the rate of change of a quantity over a closed interval is the net change of the quantity over the same interval.


Rectilinear Motion

Consider a particle that travels along a horizontal line. Here we adopt the convention that the positive direction of motion is toward the right of the origin, and denote

\(x = x(t) =\) position of the particle at time \(t\);

\(v=v(t) =\) velocity of the particle at time \(t\);

\(s=s(t) = |v(t) | = \) speed of the particle at time \(t\);

\(a = a(t) = \) acceleration of the particle at time \(t\).

 

The displacement or net distance traveled by the particle during a time interval \([a,b]\) is \(x(b) – x(a)\).

The displacement may differ from the total distance traveled by the particle over the same time interval.

Consider, for example, a particle that travels from point \(A\) to point \(B\), and then back to \(A\). Then, the net distance traveled by the particle is \(0\), while the total distance traveled is \(2|AB|\).

The Net Change Theorem implies that the displacement over a time interval \([a,b]\) is given by
\[\text{displacement}\;= \int_a^bv(t)\, dt.\]
While, the total distance traveled is given by
\[\text{total distance traveled}\; = \int_a^b|v(t)|\, dt.\]

Example 1.
Assume the velocity of a particle following a straight line trajectory is given by \(v(t)=t^2 – t -2\) m/s. Find (a) the displacement; (b) the total distance traveled by the particle over the time interval \([0,4]\).

Solution. (a)
\[\text{Displacement}\;= \int_0^4v(t)\, dt = \int_0^4(t^2 – t -2)\, dt = \frac{16}{3}\;(\text{m}).\]
Since
\[|v(t)| =| t^2 – t -2| =| (t-2)(t+1)| = \left [\renewcommand{\arraystretch}{1.2}\begin{array}{ll}
(t-2)(t+1)& \text{if} \; x\in (-\infty, -1]\cup [2,\infty)\\
-(t-2)(t+1) & \text{if}\; x\in (-1,2)
\end{array}\right .,
\]
\[\text{Distance Traveled }\;= \int_0^4|v(t)|\, dt = \int_0^2-(t^2 – t -2)\, dt + \int_2^4(t^2 – t -2)\, dt = 12\;(\text{m}).\]

Example 2.
Assume the velocity of a particle following a straight line trajectory is given by \(v(t)=\sin \pi t – \cos 2\pi t\) m/s. Find (a) the displacement; (b) the total distance traveled by the particle over the time interval \([0,4]\).

Solution. (a)
\[\text{Displacement}\;= \int_0^1v(t)\, dt = \int_0^1(\sin \pi t – \cos 2\pi t)\, dt = \frac{2}{\pi}\;(\text{m}).\]
In order to rewrite the speed as a piecewise function, we draw the graphs of \(\sin \pi t \) and \( \cos 2\pi t\) and determine the intervals where one is above the other on \([0,1]\).

Graphs of y = sin pi x and y = cos 2pi x over the interval [0,1] showing the points of intersection at x =1/6 and x=5/6

Figure 5.2

The \(x\)-coordinates of the points of intersection of the graphs on the interval \([0,1]\) are give by
\[\sin \pi x = \cos 2\pi x \Longleftrightarrow \sin \pi x = 1-2\sin^2\pi x \Longleftrightarrow (2\sin \pi x -1) (\sin \pi x+1)=0
\Longleftrightarrow x \in \left \{\frac{1}{6}, \frac{5}{6}\right \}.\]
Thus, we have
\[|v(t)| =| \sin \pi t – \cos 2\pi t| = \left [\renewcommand{\arraystretch}{2}\begin{array}{ll}
\cos 2\pi t – \sin \pi t &\displaystyle \text{if} \;t\in \left (0, \frac{\pi}{6}\right ]\cup \left [\frac{5\pi}{6},1\right ]\\
\sin \pi t – \cos 2\pi t &\displaystyle \text{if}\; t\in \left (\frac{\pi}{6}, \frac{5\pi}{6}\right )
\end{array}\right .,
\]
\[
\text{Distance Traveled }\; = \int_0^1|v(t)|\, dt = \]
\[= \int_0^{1/6}(\cos 2\pi t – \sin \pi t)\, dt +
\int_{1/6}^{5/6}(\sin \pi t – \cos 2\pi t )\, dt +
\int_{5/6}^1(\cos 2\pi t – \sin \pi t)\, dt = \frac{3\sqrt{3}-2}{\pi}\;(\text{m}).
\]

5.3.4 Exercises

In problems 14, a particle follows a straigh line trajectory with the given velocity. Find (a) the displacement; (b) the total distance traveled by the particle over the indicated time interval. Assume the units for the velocity are meters per second.

  1. \(v(t) = 2t -6\), \(t \in [1,4]\).
  2. \(v(t) =-6 t^2+t+12 \), \(t \in [ 1,3 ]\).
  3. \(v(t) =\cos \pi t – \cos 2\pi t \), \(t \in [ 1,2 ]\).
  4. \(v(t) =\cos \pi t – \sin 2\pi t \), \(t \in [ 0,1 ]\).