Week 7: Exercises

Exercises – Long Day

1: Computational Rules for Antiderivatives. By Hand

Find the indefinite integral \(\int \left( 5\cos(x+1)-\sin(5x)+\frac{2}{x-3}-7\right)\mathrm{d}x\) for \(x>3\), and explain the rules you have used along the way.

Answer

\[\begin{equation*} 5\sin(x+1)+\frac 15\cos(5x)+2\ln(x-3)-7x+C, \quad x>3, \end{equation*}\]

where \(C \in \mathbb{R}\) is an arbitrary constant.

2: Riemann Sum for a Linear Function

We are given the function \(f:[0,5] \rightarrow \mathbb{R}\) by the expression \(f(x)=2x+3\).

Question a

Find a value of the Riemann sum over the interval \([0,5]\) with 30 subintervals using Python, where the left endpoints of the subintervals are used. Then repeat for the right endpoints of the subintervals.

Question b

Find the exact values of the Riemann sum over the interval \([0,5]\) with \(n\) subintervals, where the left endpoints of the subintervals are used. Then repeat for the right endpoints of the subintervals.

Question c

Provide an exact expression of the maximum and minimum error incurred from using the above Riemann sum to calculate the area under the graph. Both expressions should only depend on \(n\).

Question d

Argue, in this specific case, that the Riemann sum has the same limit value regardless of the choice of point in the subinterval.

3: Using the Fundamental Theorem of Calculus

Question a

Provide an antiderivative of \(\frac{1}{1+x^2}\). Then calculate the integral \(\int_0^1\frac{1}{1+x^2}\mathrm{d}x\).

Hint

If you can’t remember an antiderivative of \(\frac{1}{1+x^2}\), then use SymPy to find one.

Answer

An antiderivative is \(\arctan\). The integral is \(\displaystyle{\frac{\pi}{4}}\).

Question b

Calculate the double-integrals

\[\begin{equation*} \int_1^2\Big (\int_0^1\frac{\mathrm{e}^{2x}}{y}\mathrm{d}x\Big)\mathrm{d}y \end{equation*}\]

and

\[\begin{equation*} \int_0^{\frac{\pi}{2}}\Big (\int_0^1y\cos(xy)\mathrm{d}x\Big)\mathrm{d}y. \end{equation*}\]

Hint

First calculate the inner integral while you treat \(y\) as a constant. Then do the outer integral with \(y\) treated as a variable again.

Answer

\[\begin{equation*} \frac12\left(\mathrm{e}^2-1\right)\ln(2)\quad \text{and}\quad 1 \end{equation*}\]

Question c

Let \(f: [-5,5] \to \mathbb{R}\) be given by

\[\begin{equation*} f(x) = \begin{cases} 1 & \text{for } x \in [0,1] \\ 0 & \text{for } x \in [-5,5] \setminus [0,1]. \end{cases} \end{equation*}\]

Calculate:

\[\begin{equation*} F(x) = \int_{x_{0}}^{x} f(y) \mathrm{d} y \quad \text{for $x \in [-5,5]$}, \end{equation*}\]

where \(x_{0}\in [-5,5]\) is fixed but arbitrary. You can i.e. choose \(x_0=0\). Is \(F\) continuous? Is \(F\) differentiable at all points? Is \(F\) an antiderivative of \(f\)?

Hint

Use SymPy for help if you get stuck.

4: Parametrizations in the Plane. By Hand

Consider the sets \(U \subset \mathbb{R}^2\) given by

\[\begin{equation*} U = B((0,0),2) \setminus \overline{B((0,0),1)}, \end{equation*}\]

where \(B((0,0),r)\) is an open circular disc with radius \(r\) and center \((0,0)\), and \(\overline{B((0,0),r)}\) is the closed circular disc (so, including the boundary).

Make a sketch of the set \(U\). Provide a parametrization \(\boldsymbol p:\,\,]1,2[\,\times[0,2\pi[\,\,\to U\) of \(U\). The vector function \(\boldsymbol p\) must be a function of two variables, i.e., \((r,\theta)\in\,\,]1,2[\,\times [0,2\pi[\,\), and the image set of \(\boldsymbol p\) must be \(U\).

Hint

Use polar coordinates as you remember from Mathematics 1a.

5: The Trapezoidal Method and Rieman Sums

There are many integrals that cannot be calculated exactly, often because the antiderivative cannot be expressed by a “known” function. In this exercise, we wish to calculate an approximate value for:

\[\begin{equation*} \int_0^3 \sin(x^2) \exp(3x) \mathrm{d}x. \end{equation*}\]

The SciPy package in Python can compute (approximations of) integrals using so-called numerical integration. Here, we will compare SciPy’s quad command with both the Riemann sum we know from the book and the so-called trapezoidal method.

By the trapezoidal method we mean the approximation to the integral over a small interval \([x_{j-1},x_j]\) given by

\[\begin{equation*} \int_{x_{j-1}}^{x_{j}}f(x)\mathrm{d}x=\frac{1}{2}(x_{j}-x_{j-1})(f(x_{j-1})+f(x_{j})), \end{equation*}\]

while we by mid-sums of the Riemann integral with the choice \(\xi_j := \frac{x_{j}+x_{j-1}}{2}\) have

\[\begin{equation*} \int_{x_{j-1}}^{x_{j}} f(x)dx\approx f(\xi_j) (x_{j}-x_{j-1}). \end{equation*}\]

If you want to approximate an integral over a larger interval \([a,b]\), you can subdivide it into several smaller intervals \(Q_j=[x_{j-1},x_j]\), \(j=1, \dots, J\), and approximate them individually, after which you can sum them all together – just like with Riemann sums.

Question a

Argue that \(\sin(x^2) \exp(3x)\) has an antiderivative and that the integral \(\int_0^3 \sin(x^2) \exp(3x) \, \mathrm{d}x\) is well-defined. Try (in SymPy) to find the exact value of

\[\begin{equation*} \int_0^3 \sin(x^2) \exp(3x) \mathrm{d}x. \end{equation*}\]

Use .evalf() to get an approximate value of the integral.

Answer

Solution in SymPy:

\[\begin{equation*} \int_0^3\sin(x^2)\exp(3x)\mathrm{d}x\approx 1252.4529317 \end{equation*}\]

You can (probably) get SymPy (as well as other mathematical software such as Maple) to provide an antiderivative, but it will be expressed in terms of a function \(\mathrm{erf}\), which cannot be considered “known” (to us) or to have an “explicit” definition, and therefore, it gives us no more information than the expression \(\int \sin(x^2)\exp(3x)\, \mathrm{d}x\).

Question b

Calculate the integral using quad from scipy.integrate. You must import from scipy.integrate import quad and define:

def f(x):
    return sin(x**2)*exp(3*x)

Hint

from scipy.integrate import quad

def f(x):
    return sin(x**2)*exp(3*x)

quad(f,0,3)

Answer

\[\begin{equation*} \int_0^3\sin(x^2)\exp(3x)\mathrm{d}x\approx 1252.4529317 \end{equation*}\]

Question c

We would like to compare this with the Riemann integral we have worked with earlier. We use Riemann sums, where \([a,b]\) is divided into \(J\) equal subintervals. Can you implement a function in Python that calculates this for you? It should have the following form:

def riemann_sum(f,a,b,J):

where \(f\) is a continuous function, \(a\) and \(b\) are the inteval endpoints, and \(J\) is the number of subdivisions of the integral. When you have written your Python function, you can test it on the same integral as above with \(J=20\).

Hint

It could have a structure as follows:

def riemann(f,a,b,J):
    w = (b - a)/J
    x_v = a
    result = 0
    for i in range(J):
        x_h = x_v + w
        result += ???
        x_v = x_h
    result *= w
    return result

You decide how you want to choose \(\xi_j\). Above, we have suggested choosing \(\xi_j\) as the midpoint of each subinterval.

Answer

It can look as follows:

def riemann(f,a,b,J):
    w = (b - a)/J
    x_v = a
    result = 0
    for i in range(J):
        x_h = x_v + w
        result += f((x_h+x_v)/2)
        x_v = x_h
    result *= w
    return result

Answer

\[\begin{equation*} \int_0^3\sin(x^2)\exp(3x)\mathrm{d}x\approx 1284.01 \end{equation*}\]

Question d

With the trapezoidal method we no longer approximate the area under a graph with rectangles. Can you figure out what the shape looks like based on the formula above?

Hint

See the figures on https://en.wikipedia.org/wiki/Trapezoidal_rule.

Question e

Now implement the trapezoidal method:

def trapez_sum(f,a,b,J):

where \(f\) is a continuous function, \(a\) and \(b\) are the interval endpoints, and \(J\) is the number of subdivisions of the integral. Once you’ve written your code, you can test it on the same integral as above with \(J=20\).

Question f

It doesn’t immediately seem like we get the same value of the integral. Compare your results from questions a, b, c and e. Which method is the best? Why do you think so? Also, try both with more and fewer subdivisions of the interval.

6: An Indefinite Integral. By Hand

Question a

The following is not a Rieman integral:

\[\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d} x.\]

Why not?

Answer

The integral is not (at first glance) well-defined since \(\frac{1}{\sqrt{x}}\) is not defined on \([0,1]\) (and therefore not continuous on \([0,1]\)). However, the integrand \(\frac{1}{\sqrt{x}}\) is well-defined and continuous on \([a,1]\) for all values of \(0<a<1\), which we will use in the next question.

Question b

Calculate \(\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d} x\) as \(\lim_{a \to 0} \int_a^1 \frac{1}{\sqrt{x}} \mathrm{d} x\).

Hint

Find the integral \(\int_a^1 \frac{1}{\sqrt{x}} \mathrm{d} x\) and consider the limit for \(a \to 0\).

Answer

\(\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d} x = \lim_{a \to 0} [2 \sqrt{x}]_a^1 = \lim_{a \to 0} ( 2-2\sqrt{a}) = 2\)

Question c

Find in a similar fashion the integral \(\int_1^\infty \frac{1}{\sqrt{x}} \mathrm{d} x\) (if possible).

Hint

Find the integral \(\int_1^b \frac{1}{\sqrt{x}} \mathrm{d} x\) and consider the limit for \(b \to \infty\).

Answer

The integral does not converge: \(\int_1^b \frac{1}{\sqrt{x}} \mathrm{d} x = [2 \sqrt{x}]_1^b = = 2\sqrt{b} - 2 \to \infty\) når \(b \to \infty\).

7: Change of Variables for Integrals in 2D

Determine the integral

\[\begin{equation*} \int_{0}^1\int_1^8\frac{y}{x+xy}\mathrm{d} x\mathrm{d}y \end{equation*}\]

using change of variables.

Hint

Factorize \(1/x\) out of the fraction.

Answer

\[\begin{equation*} \int_1^8\frac{y}{x+xy}\mathrm{d} x=\frac{y}{y+1}\ln(8) \end{equation*}\]

Hint

You might want to change variables as this: \(z=y+1\).

Hint

After the variable change, the indefinite integral can be written as

\[\begin{equation*} \int \frac{z-1}{z}\mathrm{d}z = \int 1 \mathrm{d}z - \int 1/z \mathrm{d}z. \end{equation*}\]

Answer

\[\begin{equation*} \int_{0}^1\int_1^8\frac{y}{x+xy}\mathrm{d} x\mathrm{d}y=\ln(8)(1-\ln(2)) \end{equation*}\]

Exercises – Short Day

1: Indefinite and Definite Integrals. By Hand

Question a

Determine an antiderivative of each of the functions

\[\begin{equation*} x^3, \quad \frac{1}{x^3} \quad \text{and} \quad \sin(3x-\frac{\pi}{2}). \end{equation*}\]

Hint

Note that the first example was treated in exercise I: Antiderivatives for Memorization.

Hint

Function no. 2: use the rewriting \(\displaystyle{\frac{1}{x^3}=x^{-3}}\).

Answer

\[\begin{equation*} \frac{1}{4}x^4,-\frac{1}{2x^2}\mathrm{and}-\frac{1}{3}\cos(3x-\frac{\pi}{2}) \end{equation*}\]

Question b

Calculate the following Riemann integrals:

\[\begin{equation*} \int_0^{1}x^3\mathrm{d}x, \quad \int_1^{2}\frac{1}{x^3}\mathrm{d}x \quad \text{and} \quad \int_{-\frac{\pi}{2}}^{0}\sin(3x-\frac{\pi}{2})\mathrm{d}x. \end{equation*}\]

Answer

\(\frac{1}{4}, \frac{3}{8}\) and \(\frac{1}{3}\).

2: Partial Integration. By Hand

Question a

We will first prove the formula for partial integration. Begin by differentiating the expression on the right-hand side of

\[\begin{equation*} \int f(x)g(x)\mathrm{d}x = F(x)g(x)- \int F(x)g^\prime(x) \mathrm{d} x. \end{equation*}\]

Hint

Remember how we differentiate the product of two functions. This is known as the product rule.

Hint

Remember that if we differentiate \(\int h(x) \mathrm{d} x\) then by definition we get \(h(x)\).

Now finish the proof.

Question b

Determine an antiderivative of the function \(x\cos(x)\), and check that it is correct.

Hint

Use partial integration, where you let \(f(x)=\cos(x)\) and \(g(x)=x\).

Hint

See examples in the book.

Answer

\(\cos(x)+x\sin(x)\). By doing \((\cos(x)+x\sin(x))'=-\sin(x)+\sin(x)+x\cos(x)=x\cos(x)\), we can confirm that the this indeed is an antiderivative.

Question c

Find an antiderivative of \(\ln(x)\) using partial integration.

Hint

You can always multiply \(\ln(x)\) by \(1\) without changing the integrand.

Hint

Rewrite \(\ln(x)\) to \(f(x)\, \ln(x)\), where \(f(x)=1\).

3: Integration by Substitution. By Hand

For the questions in this exercise we will use the method known as integration by substitution:

\[\begin{equation*} \int{f(g(x))g'(x)\mathrm{d}x}=\int{f(t)\mathrm{d}t}\quad \text{where } t=g(x). \end{equation*}\]

Question a

Determine an antiderivative of \(\displaystyle{x\mathrm{e}^{x^2}}\).

Hint

Can you identify an inner function \(g(x)\) and an outer function \(f(x)\)?

Hint

If we let \(g(x)=x^2\) and \(f(x)=\mathrm{e}^x,\) we have \(f(g(x)) g'(x)=\mathrm{e}^{x^2} 2x=2(x\mathrm{e}^{x^2})\).

Answer

\(\displaystyle{\int{x\mathrm{e}^{x^2}\mathrm{d}x}=\frac 12\int{f(t)\mathrm{d}t}=\frac 12\mathrm{e}^t=\frac 12\mathrm{e}^{x^2}}\).

Question b

Find the indefinite integral \(\displaystyle{\int \frac{x}{x^2+1} \mathrm{d}x}\).

Hint

Can you identify an inner function \(g(x)\) and an outer function \(f(x)\)?

Hint

If we let \(g(x)=x^2+1\) and \(f(x)=\displaystyle{\frac{1}{x}},\) we have \(f(g(x))g'(x)=2\frac{x}{x^2+1}\).

Answer

\[\begin{equation*} \int \frac{x}{x^2+1} \mathrm{d}x=\frac 12\int f(t)\mathrm{d}t=\frac 12\int {\frac{1}{t}}\mathrm{d}t=\frac 12\ln(t)+C=\frac 12\ln(x^2+1)+C. \end{equation*}\]

Question c

Find an antiderivative of \(\displaystyle{\frac{\sin (x)}{3 -\cos(x)}}\), and then calculate \(\displaystyle{\int_0^{\pi} \frac{\sin (x)}{3 -\cos(x)} \mathrm{d}x}\).

Hint

Set \(g(x)\) equal to the denominator, and so on.

Answer

\(\displaystyle{\int_0^{\pi} \frac{\sin (x)}{3 -\cos(x)} dx = \ln(4)-\ln(2)=\ln(2)}\).

4: Sequences. By Hand

In this and the following exercises, we are given samples of an important building block for integral calculus: sequences and their potential convergence. From Den Store Danske (Gyldendal dictionaries), translated:

Convergence, a concept of fundamental importance in mathematical analysis, especially in the theory of infinite series. A sequence of real numbers \(x_1, x_2, \ldots\) is called convergent if there exists a number \(x\) such that the number \(x_n\) is arbitrarily close to \(x\), as long as \(n\) is sufficiently large (\(\ldots\)). The number \(x\) is called the limit (or limit value) of the sequence, which is said to converge to \(x\). If the sequence is not convergent, it is called divergent.

More precisely, a sequence \(x_1, x_2, \ldots\) is said to be convergent if there exists a number \(x\) with the following property:

\[\begin{equation*} \forall \epsilon >0 \exists N \in \mathbb{N}: n \ge N \Rightarrow |x_n -x| < \epsilon. \end{equation*}\]

Four sequences \((a_n)_{n=1}^\infty\), \((b_n)_{n=1}^\infty\), \((c_n)_{n=1}^\infty\) and \((d_n)_{n=1}^\infty\) are given by

\[\begin{equation*} a_n=\frac 1n, \: b_n=\frac{n-1}{2n}, \: c_n=\frac{n}{1000} \:\text{and}\; d_n=\frac{4n^2+16}{8-3n^2} \end{equation*}\]

for \(n \in \mathbb{N}\). A sequence is written in short as \((a_n)\) for \((a_1, a_2, \dots)\) and can be considered an infinite ordered list.

Determine which of the four sequences are convergent. Specify the limit of those that are convergent.

Note

The concept of convergence is not only important in mathematical analysis. It is also the precise description of “engineering statements” such as:

Our algorithm/method/etc. converges if we simply include enough measurement points/data points/samples/etc.

Hint

If you can’t prove it from the definition, try using Python to experiment with it. You can quickly calculate the first 100 or 1000 numbers in the sequence.

Answer

\((a_n)\) is convergent with the limit 0. \((b_n)\) is convergent with the limit \(\frac 12\). \((c_n)\) is divergent. \((d_n)\) is convergent with the limit \(-\frac 43\).

5: Integrals via Left-Sums. By Hand

We will calculate the Riemann integral \(\displaystyle{\int_0^1 f(x) \mathrm{d}x}\) of the function

\[\begin{equation*} f(x)=x, \quad x\in \left[0,1\right] \end{equation*}\]

directly from the definition (so, do not find an antiderivative \(F(x)=x^2/2\) and then use it to calculate \(F(1)-F(0)=1/2-0=1/2\)).

We subdivide the interval \([0,1]\) into \(n\) equally large pieces, i.e., \(x_j = j/n\) for \(j = 0, 1, 2, \dots, n\). The Riemann sum \(S_n\) is called a left sum \(V_n\) if we always evaluate \(f\) at the left endpoint of each subinterval, that is, \(\xi_j = x_{j-1}\) for \(\xi_j \in [x_{j-1}, x_j]\) for \(j = 1, 2, \dots, n\).

Use this to determine the value of \(\displaystyle{\int_0^1 x \, \mathrm{d}x} = \lim_{n \to \infty} V_n\).

Hint

We must first find \(V_n\) expressed in terms of only \(n\).

Hint

The rectangles have areas \(\displaystyle{0,\frac{1}{n^2},\frac{2}{n^2},\frac{3}{n^2}\ldots\frac{n-1}{n^2}}\). What is the sum of these \(n\) terms?

Hint

Assume that \(n\) is odd: Ignore the first zero. Then, add the second term and the last term, add the third term and the second-to-last term, and continue in this way. In total, you will need to add two terms together \((n-1)/2\) times.

Hint

Each sum of two terms has the value \(n/n^2\), and you have \((n-1)/2\) of these. What will the final sum be?

Answer

\[\begin{equation*} V_n=\frac{n}{n^2} \frac{n-1}{2} = \frac{1}{2} \frac{n(n-1)}{n^2} \rightarrow \frac 12 \quad \text{for } n\rightarrow\infty \end{equation*}\]

So, we have

\[\begin{equation*} \int_0^1 x\mathrm{d}x=\frac 12. \end{equation*}\]