Integrals and the Fundamental Theorem of Calculus

(cod: P-67-37-4) The objective of this exercise is to obtain the area of the region \(\mathcal{R}\) bounded by the parabola \(y = x^2\) and the lines \(y = 0\) (axis \(x\)) and \(x=1\), using an exhaustion method.

Fixing a natural number \(n\), we divide the interval \([0,1]\) on the \(x\) axis into \(n\) equal subintervals and then draw vertical segments to divide the region \(\mathcal{R}\) into \(n\) strips. The figure below illustrates the situation for \(n = 5\).

First, we approximate the area of each of the \(n\) strips by the area of a rectangle, whose base lies on the \(x\) axis (measuring \(1/n\)) and whose height is given by the vertical segment coinciding with the right side of the respective strip. The figure below illustrates the situation when \(n = 5\).

We can then approximate the area of \(\mathcal{R}\) by the sum of the areas of the \(n\) rectangles, which we denote by \(S_n\). We have \[\small{\begin{eqnarray*} S_n &=& \frac{1}{n}\left(\frac{1}{n}\right)^2 + \frac{1}{n}\left(\frac{2}{n}\right)^2 + \cdots + \frac{1}{n}\left(\frac{n}{n}\right)^2 \\ & & \\ &=& \dfrac{1}{n^3}\left(1^2 + 2^2 + \cdots + n^2 \right). \end{eqnarray*}} \] Another possibility would be to approximate the area of each strip by the area of a rectangle with base \(\frac{1}{n}\) and height given by the vertical segment coinciding with the left side of the strip (note that the first rectangle will have zero height, so its area will also be zero). The figure shows the situation when \(n = 5\):

We can then approximate the area of \(\mathcal{R}\) by the sum of the areas of the \(n\) rectangles, which we denote by \(s_n\). We have \[\small{\begin{eqnarray*} s_n &=& \frac{1}{n}\left(\frac{0}{n}\right)^2 + \frac{1}{n}\left(\frac{1}{n}\right)^2 + \cdots + \frac{1}{n}\left(\frac{n-1}{n}\right)^2 \\ & & \\ &=& \dfrac{1}{n^3}\left(1^2 + 2^2 + \cdots + (n-1)^2 \right). \end{eqnarray*}} \] It is possible to verify that the approximations \(A(\mathcal{R}) \approx S_n\) and \(A(\mathcal{R}) \approx s_n\) improve as we increase the value of \(n\), as suggested by the animations below:

First, note that for each \(n \in \mathbb{N}\), we have \[s_n \leq A(\mathcal{R}) \leq S_n.\] Now, using the expressions for \(s_n\) and \(S_n\) described above, verify that \(s_n\) and \(S_n\) tend to the same real number as \(n\) tends to infinity. We then define \(A(\mathcal{R})\) as being equal to this number. Select the correct option:

(cod: P-67-38-6) The objective of this exercise is to explore the concept of definite integrals.

Consider a function \(f\) defined on an interval \([a,b]\). A partition of the interval \([a,b]\) is a division of \([a,b]\) into subintervals. More precisely, a partition is obtained by choosing points \[\small{a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b},\] such that the interval \([a,b]\) is divided into \(n\) subintervals \[ [x_0,x_1], [x_1,x_2], \dots, [x_{n-1},x_n]. \] In each subinterval of the partition, we choose a sample point. For \( i \in \{1,2,\dots,n\}\), we denote the \(i\)-th sample point by \[x_i^* \in [x_{i-1},x_i]\] and denote the length of the \(i\)-th subinterval by \[\Delta x_i = x_i - x_{i-1}\]. Given these choices, we then have a Riemann sum for \(f\) given by \[\small{f(x_1^*)\Delta x_1 + f(x_2^*)\Delta x_2 + \cdots + f(x_n^*) \Delta x_n},\] or, using sigma notation, \[\sum_{i=1}^n f(x_i^*) \Delta x_i.\] The figure below illustrates the situation with \(n = 6\). In it, the bases of the rectangles represent the subintervals of the partition (we do not display the points \(x_1, x_2, \dots, x_n\) of the partition, only the sample points).

In the example above, each of the first four terms of the Riemann sum represents the area of a blue rectangle, while the last and penultimate terms are negative (since \(f\) is negative at the last two sample points), and the value of each of these terms is obtained by multiplying the area of a pink rectangle by \((-1)\).

We say that \(f\) is integrable on the interval \([a,b]\) if there exists a number \(L\) such that \[ \lim\limits_{\text{max } \Delta x_i \to 0} \sum_{i=1}^n f(x_i^*) \Delta x_i = L,\] where this limit means that we can make a Riemann sum as close as we want to \(L\), regardless of the choice of sample points, provided that the maximum length of the subintervals of the partition is sufficiently small. In this case, we say that the definite integral of \(f\) from \(a\) to \(b\) is equal to \(L\). We use the notation \(\int_a^b f(x) \, dx\) to represent this integral, so that \[\int_a^b f(x) \, dx = \lim\limits_{\text{max } \Delta x_i \to 0} \sum_{i=1}^n f(x_i^*) \Delta x_i.\]

There is an important theorem that guarantees that if \(f\) is continuous on \([a,b]\), then \(f\) is integrable on \([a,b]\).

Given \(f\) as a continuous function on \([a,b]\) (hence integrable), we can approximate the value of the integral of \(f\) by systematically taken Riemann sums. For each \(n \in \mathbb{N}\), let us take a regular partition of \([a,b]\) into \(n\) subintervals of equal length \(\Delta x = \dfrac{b-a}{n}\). Note that \(\Delta x \longrightarrow 0\) as \(n \longrightarrow + \infty\). Thus, regardless of the choice of sample points, we have that \[\int_a^b f(x) \, dx = \lim\limits_{n \to + \infty} \sum_{i=1}^n f(x_i^*) \Delta x.\] Of course, once a regular partition is fixed, there are numerous distinct ways to choose sample points. Some common choices include, for example, taking the midpoint of each subinterval as the sample point. In the animation below, which illustrates the process as \(n\) increases, we use the right endpoint of each subinterval as the sample point. For this choice, the associated Riemann sum is called the right Riemann sum.

We observe that when the function \(f\) is non-negative on the interval \([a,b]\) (that is, \(f(x) \geq 0\) for all \(x \in [a,b]\)), the integral of \(f\) represents the area of the region between the x-axis and the graph of \(f\), with \(x\) between \(a\) and \(b\). However, this is not the case in general. In the example of our illustration, the value of the integral can be geometrically interpreted as the difference between the area of the blue region and the area of the pink region, as indicated below.

Let us now consider a specific example. Consider the function \(f(x) = x^2 - 1\) and the interval \([-1,2]\). For each \(n\), consider the right Riemann sum of \(f\) on this interval. We have that \[\Delta x = \dfrac{2 - (-1)}{n} = \dfrac{3}{n}\] and that the \(i\)-th sample point is given by \[x_i^* = x_{i+1} = -1 + i\left(\dfrac{3}{n}\right).\] Thus, we have that \[f(x_i^*) = \left(-1 + \dfrac{3i}{n}\right)^2 - 1 = \dfrac{9i^2}{n^2} - \dfrac{6i}{n}.\] Therefore, \[f(x_i^*)\Delta x = 9 \left(\dfrac{3i^2}{n^3} - \dfrac{2i}{n^2} \right).\] By working algebraically with the expression of the Riemann sum and then calculating the limit, we can conclude that

(cod: P-67-43-8) Consider an object moving with a scalar velocity \(V(t)\) (in meters per second). Recall that scalar velocity is the magnitude of the velocity vector, and therefore, it is always non-negative.

If the scalar velocity is constant over a time interval \([a,b]\), that is, if \(V(t) = V_0\), for all \(t \in [a,b]\), then the distance traveled by the object during the time interval \([a,b]\) is equal to \(V_0 (b-a)\).

Now, consider the case where the scalar velocity is not constant. Divide the interval \([a,b]\) into \(n\) subintervals of equal length \(\Delta t = \dfrac{b-a}{n}\). If we choose \(n\) large, these subintervals become very small, allowing us to assume that the velocity varies little within each subinterval. As an approximation, consider that the velocity in the \(i\)-th subinterval is constant and equal to \(V(t_i^*)\), where \(t_i^*\) is a sample point within this subinterval. We can then approximate the distance traveled in the \(i\)-th subinterval by \(V(t_i^*)\Delta t\). The distance traveled during the time interval \([a,b]\) can then be approximated by \[L \approx \sum_{i = 1}^n V(t_i^*) \Delta t.\] The exact value of the distance traveled during the time interval \([a,b]\) is then given by \[L = \lim\limits_{n \to + \infty} \sum_{i = 1}^n V(t_i^*) \Delta t = \int_a^b V(t) \, dt.\] Now consider an object launched vertically (in rectilinear motion), whose height (position), in feet, after \(t\) seconds is given by \[s = s(t) = 144t - 16t^2, \, \, t \in [0,9].\] The velocity of the object is given by \(v(t) = \dfrac{ds}{dt}\). Note that, since the motion is rectilinear, the velocity always points in the same direction (the direction of the motion axis), but not necessarily in the same sense. The scalar velocity is given by \(V(t) = |v(t)|\).

Consider the following questions: (1) Determine the distance traveled \[L = \int_0^9 V(t) \, dt.\] (2) Determine the value of the integral \[\int_a^b v(t) \, dt.\]

(cod: P-68-41-9) The objective of this exercise is to present a proof for the first part of the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus (Part 1): Let \(f\) be a continuous function on an open interval \(I\) and let \(a \in I\). The function \[F(x) = \int_a^x f(t) \, dt\] is a primitive or antiderivative of \(f\) on \(I\), that is, \[\dfrac{d}{dx}F(x) = f(x).\]

Note: Observe that the variable of the function \(F\) is the upper limit of integration. Inside the integral, to avoid confusion, we use another letter to represent the variable (called the dummy variable). Here we use \(t\), but any other letter could have been used.

Consider \(x \in I\) (in the figure below, we take \(x > a\), but the argument can be made in any case) and take \(h > 0\) such that \(x+h\) still lies in \(I\).

We have that \[F(x+h) - F(x) = \int_x^{x+h} f(t) \, dt. \] On the other hand, there exists \(c = c_h\) (which depends on \(h\)) between \(x\) and \(x+h\) such that \[\int_x^{x+h} f(t) \, dt = f(c_h)h. \] In the case where \(f\) is positive, this information means, geometrically, that for a certain \(c_h\), the area of the brown strip coincides with the area of the rectangle of height \(f(c_h)\) and base \(h\).

Therefore, \begin{eqnarray*}& & \lim\limits_{h \to 0^+} \dfrac{F(x+h) - F(x)}{h} \\ & & \\ &=& \lim\limits_{h \to 0^+} f(c_h) = f(x), \end{eqnarray*} since \(c_h \longrightarrow x\) when \(h \to 0^+\) and \(f\) is continuous on \(x\). We can use the same argument for the other lateral limit, which then allows us to conclude that \[F'(x) = \lim\limits_{h \to 0} \dfrac{F(x+h) - F(x)}{h} = f(x),\]

as we wanted to demonstrate.

Note: Part 1 of the Fundamental Theorem of Calculus states that any continuous function on an open interval has a primitive. Moreover, the theorem provides an expression for such a primitive through an integral.

Note: According to the theorem, we can write \[ \dfrac{d}{dx} \int_a^x f(t) \, dt = f(x).\]

Note: If \(f: [a,b] \to \mathbb{R}\) is continuous on the compact interval \([a,b]\) and \(F(x) = \int_a^x f(t) \, dt\), the same argument proves that \(F\) is differentiable on \((a,b)\) (with \(F'(x) = f(x)\)) and has a right-hand derivative at \(a\) and a left-hand derivative at \(b\), thus being continuous on \([a,b]\).

We then have some questions?

(1) Why is the statement in brown valid?

(2) What result guarantees the validity of the phrase in blue?

(3) What is the derivative of the function below? \[F(x) = \int_0^x e^{-t^2} \, dt\]

(cod: P-68-40-3) Fundamental Theorem of Calculus: If \(f(x)\) is a continuous function on an interval \([a,b]\) and \(F\) is a primitive or antiderivative of \(f\), that is, \[\dfrac{d}{dx}F(x)=f(x),\] then \[\int_a^b f(x) \, dx = F(b) - F(a).\]

The objective of this exercise is to present a proof for the Fundamental Theorem of Calculus (F.T.C.). For each natural \(n > 3\), consider the regular partition \[\small{a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b}\] of \([a,b]\) into subintervals of length \(\Delta x = (b-a)/n\). Note that we can express the variation of \(F\) on \([a,b]\) as the sum of the variations of \(F\) on the subintervals. In fact, we have \begin{eqnarray*} & & F(b) - F(a) \\ & & \\ &=& F(x_n) - F(x_0) \\ & & \\ &=& \small{(F(x_n)- F(x_{n-1})) + (F(x_{n-1}) - F(x_{n-2}))} \\ & & \\ &+& \small{\cdots + (F(x_{2}) - F(x_{1})) + (F(x_1)-F(x_0))}. \end{eqnarray*} Indeed, observe that, in the long sum, we can cancel several terms, leaving only the first and the last (telescoping sum). Thus, we can conclude that \(\small{\begin{equation}\label{equa1}F(b) - F(a) = \sum_{i = 1}^n F(x_i) - F(x_{i-1}).\end{equation}}\) Now, for each \(i \in \{1,2,\dots,n\}\), there exists \(x_i^* \in (x_{i-1},x_i)\) such that \[\dfrac{F(x_i) - F(x_{i-1})}{x_i - x_{i-1}} = F'(x_i^*),\] or, equivalently, \[\small{F(x_i) - F(x_{i-1}) = F'(x_i^*) \Delta x = f(x_i^*) \Delta x}.\] Therefore, it follows from equation \((\ref{equa1})\) that, for any \(n > 3\), \[F(b) - F(a) = \sum_{i = 1}^n f(x_i^*) \Delta x.\] Taking \(n \to + \infty\) (always considering \(x_i^*\) as mentioned), we conclude that \[F(b) - F(a) = \int_a^b f(x) \, dx,\] as we wanted to demonstrate.

We have two questions:

(1) Which classical theorem guarantees that the statement in brown is true?

(2) Why is the statement in blue true?

(cod: P-69-44-5) The Fundamental Theorem of Calculus, in its two parts, broadly states that differentiation and integration are inverse processes. In particular, the concept of primitive or antiderivative is central to the study of Differential and Integral Calculus. We say that a function \(F(x)\) is a primitive of \(f(x)\) on an interval \(I\) if \(F'(x) = f(x)\) for all \(x \in I\). Motivated by the Fundamental Theorem of Calculus, we introduce the concept of the indefinite integral of a function \(f\), which represents the general form of a primitive of \(f\) on an interval.

Observe that if two functions \(F_1\) and \(F_2\) are primitives of the same function \(f\) on an open interval \(I\), then \(F_1\) and \(F_2\) have identical derivatives on \(I\) and therefore differ by a constant. Thus, if \(F(x)\) is a primitive of \(f(x)\) on an open interval \(I\), the indefinite integral of \(f\) is defined by \[\int f(x) \, dx = F(x) + C,\] which tells us that the general form of a primitive of \(f\) on an interval \(I\) is given by the function \(F\) plus a constant. Note that the symbol for the indefinite integral is the same as that introduced for the definite integral, except that it does not include the integration limits. Observe that the definite integral of a function on an interval \([a,b]\) is a number, while the indefinite integral represents a family of functions, or the general form of a primitive of \(f\) on an interval.

For example, we know that \(\dfrac{d}{dx}\sin(x) = \cos(x)\). We then have that \[\int \cos(x) \, dx = \sin(x) + C.\] Consider the following statements.

(1) If \(p \ne -1\), then \[\displaystyle \int x^p \, dx = \dfrac{x^{p+1}}{p+1} + C.\]

(2) \begin{eqnarray*} & & \int (x^3 - 2x^2 + 4x -2) \, dx \\ & & \\ &=& \dfrac{x^4}{4} - \dfrac{2x^3}{3} + 2x^2 - 2x + C. \end{eqnarray*}



(3) If \[\small{\begin{eqnarray*}p(x) &=& c_n x^n + c_{n-1}x^{n-1}+ \cdots + c_1x + c_0 \\ & & \\ &=& \sum_{k = 0}^n c_k x^k \end{eqnarray*}}\] is a polynomial function, then \begin{eqnarray*} & & \int p(x) \, dx \\ & & \\ &=& \small{\dfrac{c_n x^{n+1}}{n+1} + \cdots + \dfrac{c_1 x^2}{2} + c_0x + C} \\ & & \\ &=& C + \sum_{k=0}^n \dfrac{c_k x^{k+1}}{k+1}. \end{eqnarray*}

(4) \(\displaystyle \int \dfrac{1}{x} \, dx = \ln|x| + C.\)

(5) \(\displaystyle \int \sin(x) \, dx = - \cos(x) + C\).

(6) \(\displaystyle \int \cos(x) \, dx = \sin(x) + C\).

(7) \(\displaystyle \int \sec^2(x) \, dx = \tan(x) + C\).

(8) If \(0 < b \ne 1\), \(\displaystyle \int b^x \, dx = \dfrac{b^x}{\ln{b}} + C\).<%9%12%>
(9) \(\displaystyle \int e^x \, dx = e^x + C\).

(10) \(\displaystyle \int \text{sinh}\, x \, dx = \cosh{x} + C\).

(11) \(\displaystyle \int \cosh{x} \, dx = \text{sinh} \, {x} + C\).

(12) \(\displaystyle \int \dfrac{1}{1+x^2} \, dx = \arctan{x} + C\).

(13) \(\displaystyle \int \dfrac{1}{\sqrt{1 + t^2}} \, dt = \text{arsinh}\, {t} + C\).

Regarding these statements, we have that: