MT2176 Further Calculus: The Riemann Integral & FTC

The Riemann integral is defined using the concepts of partitions, lower sums, and upper sums.

1. A **partition** \(P\) of \([a, b]\) is a finite set of points \(x_0, x_1, ..., x_n\) such that \(a = x_0 < x_1 < ... < x_n = b\).

2. The **lower sum** \(L(P)\) is \(\sum_{i=1}^{n} m_i (x_i - x_{i-1})\), where \(m_i\) is the infimum of \(f(x)\) on \([x_{i-1}, x_i]\).

3. The **upper sum** \(U(P)\) is \(\sum_{i=1}^{n} M_i (x_i - x_{i-1})\), where \(M_i\) is the supremum of \(f(x)\) on \([x_{i-1}, x_i]\).

The Riemann integral \(\int_{a}^{b} f(x) dx\) is the unique number \(I\) that satisfies \(L(P) \le I \le U(P)\) for all partitions \(P\).

Source: Based on concepts from Ostaszewski, A. (1991). Advanced mathematical methods. Chapter 17.

If \(f\) is a continuous function on \([a, b]\), then the function \(F\) defined by \(F(x) = \int_{a}^{x} f(t) dt\) for \(x\) in \([a, b]\) is continuous on \([a, b]\) and differentiable on \((a, b)\), and its derivative is \(f(x)\).

\(F\\' (x) = \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)\)

This shows that differentiation and integration are inverse processes.

Source: Based on concepts from MT2176 Subject Guide and Ostaszewski, A. (1991).

Let \(F(u) = \int_{1}^{u} \cos(t) dt\). By FTC Part 1, \(F\\' (u) = \cos(u)\).

The given integral is \(g(x) = F(x^2)\). Using the chain rule:

\(g\\' (x) = F\\' (x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x\)

So, the derivative is \(2x \cos(x^2)\).

Source: Application of concepts from Wrede, R. C., & Spiegel, M. R. (2010).

If \(f\) is a continuous function on \([a, b]\) and \(F\) is any antiderivative of \(f\) (i.e., \(F\\' (x) = f(x)\)), then:

\(\int_{a}^{b} f(x) dx = F(b) - F(a)\)

This provides a powerful method for evaluating definite integrals.

Source: Based on concepts from MT2176 Subject Guide and Ostaszewski, A. (1991).

An antiderivative of \(f(x) = \sin(x)\) is \(F(x) = -\cos(x)\).

Applying FTC Part 2:

\(\int_{0}^{\pi} \sin(x) dx = F(\pi) - F(0) = -\cos(\pi) - (-\cos(0))\)

\(= -(-1) - (-1) = 1 + 1 = 2\)

Source: Application of concepts from Wrede, R. C., & Spiegel, M. R. (2010).

If a function is continuous on a closed interval \([a, b]\), then it is Riemann integrable on that interval.

However, a function does not need to be continuous to be Riemann integrable. A function is Riemann integrable if and only if it is bounded and the set of its discontinuities has measure zero (e.g., a finite number of jump discontinuities).

Source: Based on concepts from Ostaszewski, A. (1991), Chapter 17.

We write \(h(x) = \int_{x^2}^{c} e^{t^2} dt + \int_{c}^{x^3} e^{t^2} dt = -\int_{c}^{x^2} e^{t^2} dt + \int_{c}^{x^3} e^{t^2} dt\).

Differentiating using the FTC Part 1 and the chain rule:

\(h\\' (x) = -e^{(x^2)^2} \cdot (2x) + e^{(x^3)^2} \cdot (3x^2)\)

\(= 3x^2 e^{x^6} - 2x e^{x^4}\)

Source: Application of concepts from MT2176 Subject Guide.

Intersection points: \(x^2 = \sqrt{x} \Rightarrow x^4 = x \Rightarrow x(x^3 - 1) = 0\), so \(x=0, x=1\).

In \([0, 1]\), \(\sqrt{x} \ge x^2\). The area \(A\) is:

\(A = \int_{0}^{1} (\sqrt{x} - x^2) dx = [\frac{2}{3}x^{3/2} - \frac{1}{3}x^3]_{0}^{1}\)

\(= (\frac{2}{3} - \frac{1}{3}) - 0 = \frac{1}{3}\)

Source: Application of concepts from Wrede, R. C., & Spiegel, M. R. (2010).

If \(f\) is continuous on \([a, b]\), then there exists a number \(c\) in \([a, b]\) such that:

\(\int_{a}^{b} f(x) dx = f(c)(b - a)\)

The value \(f(c)\) is the average value of the function on the interval.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

A **definite integral**, \(\int_{a}^{b} f(x) dx\), is a number representing the net area under the curve of \(f(x)\) from \(a\) to \(b\).

An **indefinite integral**, \(\int f(x) dx\), is a family of functions, \(F(x) + C\), representing all antiderivatives of \(f(x)\).

Source: Based on concepts from Ostaszewski, A. (1991).

A partition \(P\) of an interval \([a, b]\) is a finite, ordered set of points \(x_0, x_1, ..., x_n\) such that \(a = x_0 < x_1 < ... < x_n = b\).

Source: Ostaszewski, A. (1991).

The lower sum is \(L(P, f) = \sum_{i=1}^{n} m_i (x_i - x_{i-1})\), where \(m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}\).

Source: Ostaszewski, A. (1991).

The upper sum is \(U(P, f) = \sum_{i=1}^{n} M_i (x_i - x_{i-1})\), where \(M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}\).

Source: Ostaszewski, A. (1991).

A Riemann sum for a function \(f\) on an interval \([a,b]\) is a sum of the form \(\sum_{i=1}^{n} f(t_i) \Delta x_i\), where \(P = \{x_0, ..., x_n\}\) is a partition of \([a,b]\), \(\Delta x_i = x_i - x_{i-1}\), and \(t_i\) is any sample point in the subinterval \([x_{i-1}, x_i]\).

Source: Ostaszewski, A. (1991).

A function \(f\) is Riemann integrable on \([a,b]\) if the lower integral and the upper integral are equal. This is equivalent to saying that for any \(\epsilon > 0\), there exists a partition \(P\) such that \(U(P, f) - L(P, f) < \epsilon\).

Source: Ostaszewski, A. (1991).

An antiderivative is \(F(x) = x^3 - x^2 + x\).

\(\int_{0}^{2} (3x^2 - 2x + 1) dx = [x^3 - x^2 + x]_{0}^{2}\)

\(= (2^3 - 2^2 + 2) - (0) = 8 - 4 + 2 = 6\)

Source: Application of FTC Part 2.

The average value is \(\frac{1}{3-0} \int_{0}^{3} x^2 dx\).

\(\frac{1}{3} [\frac{x^3}{3}]_{0}^{3} = \frac{1}{3} (\frac{3^3}{3} - 0) = \frac{1}{3} (9) = 3\)

Source: Application of Mean Value Theorem for Integrals.

First, reverse the limits: \(-\frac{d}{dx} \int_{1}^{\sin x} \ln(t) dt\).

Using FTC Part 1 and the chain rule:

\(= -\ln(\sin x) \cdot \cos x\)

Source: Application of FTC Part 1 and Chain Rule.

For a non-negative function \(f(x)\), the definite integral \(\int_{a}^{b} f(x) dx\) represents the area of the region under the curve of \(y=f(x)\) from \(x=a\) to \(x=b\).

For a general function, it represents the net area (area above the x-axis minus the area below the x-axis).

Source: MT2176 Subject Guide.

False. A classic counterexample is the Dirichlet function, which is 1 for rational numbers and 0 for irrational numbers. It is bounded on \([0,1]\), but for any partition, the lower sum is 0 and the upper sum is 1, so it is not Riemann integrable.

Source: Ostaszewski, A. (1991).

If \(f\) is integrable on an interval containing points \(a, b, c\), then:

\(\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx\)

This holds regardless of the order of \(a, b, c\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

An antiderivative of \(f(x) = 1/x\) is \(F(x) = \ln|x|\).

\(\int_{1}^{e} \frac{1}{x} dx = [\ln|x|]_{1}^{e} = \ln(e) - \ln(1) = 1 - 0 = 1\)

Source: Application of FTC Part 2.

The area is given by the integral \(\int_{0}^{1} e^x dx\).

\(\int_{0}^{1} e^x dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1\)

Source: Application of definite integrals.

Using the linearity property of integrals:

\(\int_{0}^{5} (2f(x) - 3g(x)) dx = 2\int_{0}^{5} f(x) dx - 3\int_{0}^{5} g(x) dx\)

\(= 2(10) - 3(3) = 20 - 9 = 11\)

Source: Properties of definite integrals.

A partition \(P^*\) is a refinement of a partition \(P\) if \(P \subseteq P^*\). In other words, \(P^*\) contains all the points of \(P\) and possibly some more. Adding points to a partition refines it.

Source: Ostaszewski, A. (1991).

If \(P^*\) is a refinement of \(P\), then the lower sum increases and the upper sum decreases:

\(L(P, f) \le L(P^*, f)\) and \(U(P^*, f) \le U(P, f)\)

The sums get closer to the true value of the integral.

Source: Ostaszewski, A. (1991).

The function \(f(x) = 0\) for \(x < 1/2\) and \(f(x) = 1\) for \(x \ge 1/2\) has a single jump discontinuity at \(x=1/2\). Since it is bounded and has only a finite number of discontinuities, it is Riemann integrable on \([0, 1]\).

Source: MT2176 Subject Guide.

According to the Fundamental Theorem of Calculus, Part 1, if \(F(x) = \int_{a}^{x} f(t) dt\), then \(F\\' (x) = f(x)\).

Here, \(f(t) = \sqrt{1+t^2}\). So, the derivative is simply \(\sqrt{1+x^2}\).

Source: Application of FTC Part 1.

An antiderivative of \(f(x) = \sec^2(x)\) is \(F(x) = \tan(x)\).

\(\int_{0}^{\pi/4} \sec^2(x) dx = [\tan(x)]_{0}^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1\)

Source: Application of FTC Part 2.

The integral is 0. The area from \(-a\) to 0 cancels out the area from 0 to \(a\). For example, \(\int_{-\pi}^{\pi} \sin(x) dx = 0\).

Source: Properties of definite integrals.

The integral over the symmetric interval is twice the integral over the half-interval:

\(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\)

For example, \(\int_{-1}^{1} x^2 dx = 2 \int_{0}^{1} x^2 dx\).

Source: Properties of definite integrals.

This is a Riemann sum for \(f(x) = x^4\) on the interval \([0, 1]\) with \(\Delta x = 1/n\) and sample points \(x_i = i/n\).

The limit is equal to the definite integral \(\int_{0}^{1} x^4 dx\).

Source: Definition of the definite integral.

It generalizes the Riemann integral to allow for integration with respect to a function \(\alpha(x)\) other than \(x\). This is crucial in physics (for mass distributions that are not uniform) and probability theory (for integrating with respect to cumulative distribution functions, including discrete ones).

Source: Ostaszewski, A. (1991).

In \([0, 1]\), \(x \ge x^2\). The area is:

\(A = \int_{0}^{1} (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1}\)

\(= (\frac{1}{2} - \frac{1}{3}) - 0 = \frac{1}{6}\)

Source: Application of definite integrals.

No. A function must be bounded on the interval to be Riemann integrable. The function \(f(x) = 1/x\) is unbounded at \(x=0\), which is inside the interval \([-1, 1]\). This is an improper integral.

Source: MT2176 Subject Guide.

Yes. The function \(f(x) = 1/x\) is continuous on the closed interval \([1, 2]\). Since continuity implies integrability, it is Riemann integrable on this interval.

Source: MT2176 Subject Guide.

The integral is bounded as well:

\(m(b-a) \le \int_{a}^{b} f(x) dx \le M(b-a)\)

This follows from the properties of upper and lower sums.

Source: Ostaszewski, A. (1991).

First, reverse the limits: \(-\frac{d}{dx} \int_{5}^{x} (t + \sin t) dt\).

By FTC Part 1, the derivative is \(-(x + \sin x)\).

Source: Application of FTC Part 1.

Let \(u = x+1\), so \(du = dx\). When \(x=0, u=1\). When \(x=1, u=2\).

The integral becomes \(\int_{1}^{2} u^5 du = [\frac{u^6}{6}]_{1}^{2}\)

\(= \frac{2^6}{6} - \frac{1^6}{6} = \frac{64 - 1}{6} = \frac{63}{6} = \frac{21}{2}\)

Source: Application of FTC Part 2 with substitution.

A step function is constant on a series of subintervals. The integral is the sum of the areas of the rectangles formed by the function on each subinterval. If \(f(x) = c_i\) on \((x_{i-1}, x_i))\), the integral is \(\sum_{i=1}^{n} c_i (x_i - x_{i-1})\).

Source: Ostaszewski, A. (1991).

We must use the Product Rule and FTC Part 1.

\(G\\' (x) = (1) \cdot \int_{2}^{x} \sin(t^2) dt + x \cdot \frac{d}{dx} \int_{2}^{x} \sin(t^2) dt\)

\(= \int_{2}^{x} \sin(t^2) dt + x \sin(x^2)\)

Source: Application of Product Rule and FTC Part 1.

The error \(E_T\) in approximating \(\int_a^b f(x) dx\) with the Trapezoidal Rule using \(n\) subintervals is bounded by:

\(|E_T| \le \frac{K(b-a)^3}{12n^2}\)

where \(K\) is an upper bound for \(|f\\' (x)|\) on \([a, b]\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The error \(E_S\) in approximating \(\int_a^b f(x) dx\) with Simpson's Rule using \(n\) subintervals (where \(n\) is even) is bounded by:

\(|E_S| \le \frac{K(b-a)^5}{180n^4}\)

where \(K\) is an upper bound for \(|f^{(4)}(x)|\) (the fourth derivative) on \([a, b]\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The Trapezoidal Rule approximates the function with straight lines, while Simpson's Rule uses parabolas. Parabolas can more closely follow the curvature of the function, leading to a smaller error. This is also reflected in the error bound formulas, where the error for Simpson's Rule decreases with \(n^4\) compared to \(n^2\) for the Trapezoidal Rule.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

This is an indefinite integral. We use substitution. Let \(u = 1+x^2\), so \(du = 2x dx\).

The integral becomes \(\int \frac{1}{u} du = \ln|u| + C\).

Substituting back, we get \(\ln(1+x^2) + C\). (Absolute value is not needed as \(1+x^2\) is always positive).

Source: Application of integration techniques.

A function \(f\) has bounded variation on \([a,b]\) if the total variation, \(V_a^b(f) = \sup_P \sum_{i=1}^n |f(x_i) - f(x_{i-1})|\), is finite. The supremum is taken over all partitions \(P\) of \([a,b]\). Intuitively, it means the function does not "wiggle" infinitely much.

Source: Ostaszewski, A. (1991).

A sufficient condition for the integral to exist is that \(f\) is continuous and \(\alpha\) is of bounded variation on \([a,b]\).

Source: Ostaszewski, A. (1991).

If \(f\) is continuous, the integral simplifies to a sum:

\(\int_a^b f(x) d\alpha(x) = \sum_k f(x_k) s_k\)

where the sum is over all jump points \(x_k\) in \((a,b)\).

Source: Ostaszewski, A. (1991).

One petal is traced as \(\theta\) goes from \(-\pi/4\) to \(\pi/4\). The area in polar coordinates is \(A = \frac{1}{2} \int_a^b r^2 d\theta\).

\(A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2} d\theta\)

\(= \frac{1}{4} [\theta + \frac{1}{4}\sin(4\theta)]_{-\pi/4}^{\pi/4} = \frac{1}{4} [(\frac{\pi}{4} + 0) - (-\frac{\pi}{4} + 0)] = \frac{\pi}{8}\)

Source: Application of integration in polar coordinates.

We use integration by parts: \(\int u dv = uv - \int v du\).

Let \(u=x\) and \(dv = e^x dx\). Then \(du = dx\) and \(v = e^x\).

\(\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C\)

Source: Application of integration by parts.

Suppose \(f(x)\) is a continuous, positive, decreasing function on \([1, \infty)\) and let \(a_n = f(n)\). Then the series \(\sum_{n=1}^{\infty} a_n\) is convergent if and only if the improper integral \(\int_{1}^{\infty} f(x) dx\) is convergent.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

We consider the integral \(\int_{1}^{\infty} \frac{1}{x^2} dx\).

\(\int_{1}^{\infty} \frac{1}{x^2} dx = \lim_{b \to \infty} [-\frac{1}{x}]_1^b = \lim_{b \to \infty} (-\frac{1}{b} - (-1)) = 1\)

Since the integral converges, the series also converges.

Source: Application of the integral test.

We use the disk method. The volume \(V\) is given by:

\(V = \pi \int_{0}^{1} [R(x)]^2 dx\), where the radius \(R(x) = x^2\).

\(V = \pi \int_{0}^{1} (x^2)^2 dx = \pi \int_{0}^{1} x^4 dx\)

\(= \pi [\frac{x^5}{5}]_0^1 = \frac{\pi}{5}\)

Source: Application of definite integrals for volume.

The improper integral is defined as the limit:

\(\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx\)

The integral is said to converge if this limit exists and is finite.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

\(\int_0^\infty e^{-x} dx = \lim_{b \to \infty} \int_0^b e^{-x} dx\)

\(= \lim_{b \to \infty} [-e^{-x}]_0^b = \lim_{b \to \infty} (-e^{-b} - (-e^0))\)

\(= \lim_{b \to \infty} (-e^{-b} + 1) = 0 + 1 = 1\)

Source: Application of improper integrals.

The improper integral \(\int_1^\infty \frac{1}{x^p} dx\) converges if \(p > 1\) and diverges if \(p \le 1\).

This is a useful benchmark for the comparison test for improper integrals.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The arc length formula is \(L = \int_a^b \sqrt{1 + (y\\' )^2} dx\).

\(y\\' = x^{1/2}\), so \((y\\' )^2 = x\).

\(L = \int_0^3 \sqrt{1+x} dx = [\frac{2}{3}(1+x)^{3/2}]_0^3\)

\(= \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3}\)

Source: Application of definite integrals for arc length.

This limit is the definition of the derivative of the integral function \(F(x) = \int_c^x f(t) dt\).

By the Fundamental Theorem of Calculus, Part 1, this limit is equal to \(f(x)\).

Source: Definition of derivative and FTC Part 1.

False. For example, \(\int_0^{2\pi} \sin(x) dx = 0\), but \(\sin(x)\) is not identically zero on \([0, 2\pi]\). The net area can be zero even if the function is not.

Source: Properties of definite integrals.

The lower integral is the supremum of all lower sums over all possible partitions: \(\underline{\int_a^b} f = \sup_P L(P,f)\).

The upper integral is the infimum of all upper sums over all possible partitions: \(\overline{\int_a^b} f = \inf_P U(P,f)\).

A function is Riemann integrable if and only if these two values are equal.

Source: Ostaszewski, A. (1991).

The subintervals are \([0, 1/2]\) and \([1/2, 1]\). Since \(f(x)=x\) is increasing:

\(m_1 = \inf\{f(x) : x \in [0, 1/2]\} = f(0) = 0\).

\(m_2 = \inf\{f(x) : x \in [1/2, 1]\} = f(1/2) = 1/2\).

\(L(P,f) = m_1(1/2 - 0) + m_2(1 - 1/2) = 0(1/2) + (1/2)(1/2) = 1/4\).

Source: Application of lower sums.

The subintervals are \([0, 1/2]\) and \([1/2, 1]\). Since \(f(x)=x\) is increasing:

\(M_1 = \sup\{f(x) : x \in [0, 1/2]\} = f(1/2) = 1/2\).

\(M_2 = \sup\{f(x) : x \in [1/2, 1]\} = f(1) = 1\).

\(U(P,f) = M_1(1/2 - 0) + M_2(1 - 1/2) = (1/2)(1/2) + 1(1/2) = 1/4 + 1/2 = 3/4\).

Source: Application of upper sums.

We need to find \(c\) such that \(f(c)(2-0) = \int_0^2 3x^2 dx\).

\(\int_0^2 3x^2 dx = [x^3]_0^2 = 8\).

So, \(3c^2 (2) = 8 \Rightarrow 6c^2 = 8 \Rightarrow c^2 = 4/3\).

\(c = 2/\sqrt{3}\). This value is in the interval \([0, 2]\).

Source: Application of MVT for Integrals.

First, expand the integral: \(f(x) = x \int_0^x g(t) dt - \int_0^x t g(t) dt\).

Now differentiate using the product rule and FTC:

\(f\\' (x) = [1 \cdot \int_0^x g(t) dt + x \cdot g(x)] - [x g(x)] = \int_0^x g(t) dt\)

Differentiate again:

\(f\\'\' (x) = g(x)\)

Source: Application of FTC and Product Rule.

If a function \(f\) is monotonic (either non-decreasing or non-increasing) on a closed interval \([a, b]\), then it is Riemann integrable on \([a, b]\). This is because a monotonic function is bounded and its set of discontinuities is at most countable, which has measure zero.

Source: Ostaszewski, A. (1991).

This is the Riemann sum for \(f(x) = \sqrt{1-x^2}\) on the interval \([0, 1]\).

The limit is \(\int_0^1 \sqrt{1-x^2} dx\).

This integral represents the area of a quarter-circle of radius 1, which is \(\frac{\pi (1)^2}{4} = \frac{\pi}{4}\).

Source: Definition of the definite integral.

The area is \(A = \int_1^{e^2} \frac{1}{x} dx\).

\(A = [\ln|x|]_1^{e^2} = \ln(e^2) - \ln(1) = 2 - 0 = 2\)

Source: Application of definite integrals.

You can conclude that \(f(x) = 0\) for all \(x\) in \([a,b]\). If \(f(c) > 0\) for some \(c\), then by continuity, \(f(x) > 0\) in a small neighborhood around \(c\), which would make the integral greater than 0. This contradicts the given information.

Source: Properties of definite integrals.

Let \(u = \sqrt{x} = x^{1/2}\). Then \(\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).

Using FTC Part 1 and the chain rule:

\(\frac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \cdot \frac{1}{2\sqrt{x}} = \frac{x}{x^2+1} \cdot \frac{1}{2\sqrt{x}} = \frac{\sqrt{x}}{2(x^2+1)}\)

Source: Application of FTC Part 1 and Chain Rule.

The net change theorem is a restatement of the Fundamental Theorem of Calculus, Part 2. It says that the definite integral of a rate of change is the total change:

\(\int_a^b F\\' (x) dx = F(b) - F(a)\)

For example, if \(v(t)\) is the velocity of an object, \(\int_a^b v(t) dt\) is the net change in position (displacement) from time \(a\) to \(b\).

Source: MT2176 Subject Guide.

Displacement is the definite integral of velocity:

\(\int_1^4 (t^2 - t - 6) dt = [\frac{t^3}{3} - \frac{t^2}{2} - 6t]_1^4\)

\(= (\frac{64}{3} - 8 - 24) - (\frac{1}{3} - \frac{1}{2} - 6) = (\frac{64}{3} - 32) - (\frac{2-3-36}{6})\)

\(= \frac{64-96}{3} - \frac{-37}{6} = -\frac{32}{3} + \frac{37}{6} = \frac{-64+37}{6} = -\frac{27}{6} = -4.5\) meters.

Source: Application of the Net Change Theorem.

Total distance is \(\int_1^4 |v(t)| dt\). We must find where \(v(t)\) is negative. \(v(t) = (t-3)(t+2) = 0\) at \(t=3, -2\). On \([1,4]\), \(v(t)\) is negative for \(t<3\) and positive for \(t>3\).

\(\int_1^3 -(t^2-t-6) dt + \int_3^4 (t^2-t-6) dt\)

\(= [-\frac{t^3}{3} + \frac{t^2}{2} + 6t]_1^3 + [\frac{t^3}{3} - \frac{t^2}{2} - 6t]_3^4\)

\(= ((-9 + 4.5 + 18) - (-1/3 + 0.5 + 6)) + ((\frac{64}{3} - 8 - 24) - (9 - 4.5 - 18))\)

\(= (13.5 - 6.166) + (-10.667 - (-13.5)) = 7.333 + 2.833 = 10.166\) meters.

Source: Application of definite integrals.

Let \(u = 1+x^2\), so \(du = 2x dx\) and \(x dx = du/2\).

When \(x=0, u=1\). When \(x=1, u=2\).

\(\int_1^2 \frac{1}{u} \frac{du}{2} = \frac{1}{2} [\ln|u|]_1^2 = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2\)

Source: Application of FTC Part 2 with substitution.

The norm of a partition \(P = \{x_0, x_1, ..., x_n\}\), denoted \(||P||\), is the length of the longest subinterval:

\(||P|| = \max\{x_i - x_{i-1} : i=1, ..., n\}\)

The definite integral can be defined as the limit of Riemann sums as the norm of the partition approaches zero.

Source: Ostaszewski, A. (1991).

We use the method of cylindrical shells. The volume \(V\) is:

\(V = 2\pi \int_a^b x f(x) dx\)

\(V = 2\pi \int_0^4 x \sqrt{x} dx = 2\pi \int_0^4 x^{3/2} dx\)

\(= 2\pi [\frac{2}{5}x^{5/2}]_0^4 = 2\pi (\frac{2}{5} (4^{5/2}) - 0) = 2\pi (\frac{2}{5} \cdot 32) = \frac{128\pi}{5}\)

Source: Application of definite integrals for volume.

In the Riemann integral, the Riemann sum is \(\sum f(t_i) \Delta x_i\), where \(\Delta x_i = x_i - x_{i-1}\) is the length of the subinterval.

In the Riemann-Stieltjes integral, the sum is \(\sum f(t_i) \Delta \alpha_i\), where \(\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1})\) is the change in a second function, the "integrator" \(\alpha(x)\), across the subinterval.

Source: Ostaszewski, A. (1991).

First, expand the integral: \(f(x) = x \int_0^x g(t) dt - \int_0^x t g(t) dt\).

Now differentiate using the product rule and FTC:

\(f\\' (x) = [1 \cdot \int_0^x g(t) dt + x \cdot g(x)] - [x g(x)] = \int_0^x g(t) dt\)

Differentiate again:

\(f\\'\' (x) = g(x)\)

Source: Application of FTC and Product Rule.

This is a step function. We break the integral up by integer values:

\(\int_{-2}^{-1} -2 dx + \int_{-1}^{0} -1 dx + \int_{0}^{1} 0 dx + \int_{1}^{2} 1 dx + \int_{2}^{3} 2 dx\)

\(= -2(1) - 1(1) + 0(1) + 1(1) + 2(1) = -2 - 1 + 0 + 1 + 2 = 0\)

Source: Integration of step functions.

Suppose \(f\) and \(g\) are continuous functions with \(f(x) \ge g(x) \ge 0\) for \(x \ge a\).

1. If \(\int_a^\infty f(x) dx\) is convergent, then \(\int_a^\infty g(x) dx\) is convergent.

2. If \(\int_a^\infty g(x) dx\) is divergent, then \(\int_a^\infty f(x) dx\) is divergent.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

We know that \(0 \le \sin^2 x \le 1\) for all \(x\).

Therefore, \(0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2}\).

We know that \(\int_1^\infty \frac{1}{x^2} dx\) converges (it's a p-integral with \(p=2>1\)).

By the comparison test, \(\int_1^\infty \frac{\sin^2 x}{x^2} dx\) also converges.

Source: Application of comparison test.

The surface area of the solid obtained by rotating the curve \(y=f(x)\) from \(x=a\) to \(x=b\) about the x-axis is given by the integral:

\(S = 2\pi \int_a^b f(x) \sqrt{1 + [f\\' (x)]^2} dx\)

assuming \(f(x) \ge 0\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

Consider a slice of water at height \(y\) with thickness \(dy\). By similar triangles, the radius \(r\) at height \(y\) is \(r/y = 5/10 \Rightarrow r = y/2\). The volume of the slice is \(dV = \pi r^2 dy = \pi (y/2)^2 dy\). The force is \(dF = \rho g dV\). The distance to pump is \((10-y)\).

Work \(W = \int_0^{10} (10-y) \rho g \pi (y/2)^2 dy = \frac{\rho g \pi}{4} \int_0^{10} (10y^2 - y^3) dy\)

\(= \frac{\rho g \pi}{4} [\frac{10y^3}{3} - \frac{y^4}{4}]_0^{10} = \frac{\rho g \pi}{4} (\frac{10000}{3} - \frac{10000}{4}) = \frac{\rho g \pi}{4} (\frac{10000}{12}) \approx 2.57 \times 10^6\) Joules.

Source: Application of definite integrals.

For an integral with a singularity at \(c\) inside \([a,b]\), the Cauchy Principal Value is \(\lim_{\epsilon \to 0^+} [\int_a^{c-\epsilon} f(x) dx + \int_{c+\epsilon}^b f(x) dx]\).

For an integral on \((-\infty, \infty)\), it is \(\lim_{R \to \infty} \int_{-R}^R f(x) dx\).

This can assign a value to some integrals that would otherwise diverge.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The integral is improper at \(x=0\). The P.V. is:

\(\lim_{\epsilon \to 0^+} [\int_{-1}^{-\epsilon} \frac{1}{x} dx + \int_{\epsilon}^1 \frac{1}{x} dx]\)

\(= \lim_{\epsilon \to 0^+} [(\ln|-\epsilon| - \ln|-1|) + (\ln|1| - \ln|\epsilon|)]\)

\(= \lim_{\epsilon \to 0^+} [\ln(\epsilon) - 0 + 0 - \ln(\epsilon)] = \lim_{\epsilon \to 0^+} 0 = 0\)

Source: Application of Cauchy Principal Value.

The Gamma function \(\Gamma(z)\) is a generalization of the factorial function to complex numbers. It is defined by an improper integral:

\(\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx\)

For a positive integer \(n\), \(\Gamma(n) = (n-1)!\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

Using the definition \(\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx\):

\(\Gamma(1) = \int_0^\infty x^{1-1} e^{-x} dx = \int_0^\infty e^{-x} dx\)

\(= [-e^{-x}]_0^\infty = \lim_{b \to \infty} (-e^{-b}) - (-e^0) = 0 - (-1) = 1\)

This is consistent with \((1-1)! = 0! = 1\).

Source: Application of the Gamma function.

The Beta function, \(B(x,y)\), is defined by the integral:

\(B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt\)

It is related to the Gamma function by \(B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The convolution of two functions \(f\) and \(g\), denoted \((f*g)(t)\), is defined by the integral:

\((f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau) d\tau\)

It is used extensively in signal processing, statistics (as the distribution of a sum of random variables), and in the solution of differential equations using Laplace transforms.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The Laplace Transform of \(f(t)\), denoted \(\mathcal{L}\{f(t)\}(s) = F(s)\), is defined by the improper integral:

\(F(s) = \int_0^\infty e^{-st} f(t) dt\)

It transforms a function of time \(t\) into a function of complex frequency \(s\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

\(\mathcal{L}\{1\}(s) = \int_0^\infty e^{-st} (1) dt = [-\frac{1}{s}e^{-st}]_0^\infty\)

\(= \lim_{b \to \infty} (-\frac{1}{s}e^{-sb}) - (-\frac{1}{s}e^0)\)

For \(s>0\), this limit is \(0 - (-\frac{1}{s})) = \frac{1}{s}\).

Source: Application of Laplace Transform.

The Fourier Transform of \(f(t)\), denoted \(\mathcal{F}\{f(t)\}(\omega) = \hat{f}(\omega)\), is defined by the integral:

\(\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt\)

It decomposes a function of time into its constituent frequencies.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

A line integral is an integral where the function to be integrated is evaluated along a curve \(C\).

The line integral of a scalar function \(f\) along a curve \(C\) parameterized by \(\mathbf{r}(t)\) from \(t=a\) to \(t=b\) is:

\(\int_C f ds = \int_a^b f(\mathbf{r}(t)) ||\mathbf{r}\' (t)|| dt\)

It generalizes the definite integral for functions of two or more variables.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

A double integral is an integral of a function of two variables, \(f(x,y)\), over a region \(R\) in the xy-plane.

\(\iint_R f(x,y) dA\)

It can be used to calculate the volume under the surface \(z=f(x,y)\) over the region \(R\).

Source: Wrede, R. C., & Spiegel, M. R. (2010).

First, integrate with respect to \(x\):

\(\int_0^1 [\frac{x^2}{2} + yx]_0^2 dy = \int_0^1 ((\frac{4}{2} + 2y) - 0) dy = \int_0^1 (2+2y) dy\)

Now, integrate with respect to \(y\):

\([2y + y^2]_0^1 = (2+1) - 0 = 3\)

Source: Application of double integrals.

Fubini's Theorem states that if \(f(x,y)\) is continuous on a rectangular region \(R = [a,b] \times [c,d]\), then the double integral can be computed as an iterated integral in either order:

\(\iint_R f(x,y) dA = \int_c^d \int_a^b f(x,y) dx dy = \int_a^b \int_c^d f(x,y) dy dx\)

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The Jacobian determinant is a factor used in changing variables in multiple integrals. If you change from variables \((x,y)\) to \((u,v)\), the differential area element changes:

\(dA = dx dy = |\frac{\partial(x,y)}{\partial(u,v)}| du dv\)

where \(|\frac{\partial(x,y)}{\partial(u,v)}|\) is the absolute value of the Jacobian determinant.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The transformation is \(x = r \cos \theta\) and \(y = r \sin \theta\).

The Jacobian determinant is:

\(\frac{\partial(x,y)}{\partial(r,\theta)} = \det \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{pmatrix} = \det \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}\)

\(= (r \cos^2 \theta) - (-r \sin^2 \theta) = r(\cos^2 \theta + \sin^2 \theta) = r\)

So, \(dx dy = r dr d\theta\).

Source: Application of the Jacobian.

The region is the upper half of the unit circle. In polar coordinates, this is \(0 \le r \le 1\) and \(0 \le \theta \le \pi\). The integrand \(x^2+y^2 = r^2\) and \(dy dx = r dr d\theta\).

\(\int_0^\pi \int_0^1 (r^2) r dr d\theta = \int_0^\pi \int_0^1 r^3 dr d\theta\)

\(= \int_0^\pi [\frac{r^4}{4}]_0^1 d\theta = \int_0^\pi \frac{1}{4} d\theta = [\frac{\theta}{4}]_0^\pi = \frac{\pi}{4}\)

Source: Application of polar coordinates in double integrals.

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the plane region \(D\) bounded by \(C\).

\(\oint_C (L dx + M dy) = \iint_D (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}) dA\)

It is a two-dimensional special case of the more general Stokes' Theorem.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

Area is \(\iint_D 1 dA\). We need to find \(L\) and \(M\) such that \(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1\). Common choices are:

1. \(L=0, M=x\)

2. \(L=-y, M=0\)

3. \(L=-y/2, M=x/2\)

This gives three formulas for area:

\(A = \oint_C x dy = -\oint_C y dx = \frac{1}{2} \oint_C (x dy - y dx)\)

Source: Application of Green's Theorem.

Stokes' Theorem relates the surface integral of the curl of a vector field \(\mathbf{F}\) over a surface \(S\) to the line integral of the vector field over its boundary curve \(\partial S\).

\(\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}\)

It is a generalization of Green's Theorem to three dimensions.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The Divergence Theorem relates the flux of a vector field \(\mathbf{F}\) through a closed surface \(S\) to the triple integral of the divergence of the field over the volume \(V\) enclosed by the surface.

\(\oiint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV\)

It relates a surface integral to a volume integral.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The divergence of a vector field at a point measures the magnitude of a source or sink at that point. If \(\nabla \cdot \mathbf{F} > 0\), there is a net "flow" out of the point (a source). If \(\nabla \cdot \mathbf{F} < 0\), there is a net "flow" into the point (a sink). If \(\nabla \cdot \mathbf{F} = 0\), the field is said to be incompressible.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

The curl of a vector field at a point measures the tendency of the field to "rotate" or "circulate" around that point. The direction of \(\nabla \times \mathbf{F}\) is the axis of rotation (by the right-hand rule), and its magnitude is the magnitude of the rotation. If \(\nabla \times \mathbf{F} = 0\), the field is said to be irrotational.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

A vector field \(\mathbf{F}\) is conservative if it is the gradient of some scalar function \(\phi\), called the scalar potential: \(\mathbf{F} = \nabla \phi\).

An equivalent condition for a simply-connected domain is that its curl is zero: \(\nabla \times \mathbf{F} = 0\).

Line integrals of conservative fields are path-independent.

Source: Wrede, R. C., & Spiegel, M. R. (2010).

If \(\mathbf{F}\) is a conservative vector field with potential function \(\phi\) (i.e., \(\mathbf{F} = \nabla \phi\)), then the line integral of \(\mathbf{F}\) along a curve \(C\) from point \(A\) to point \(B\) is:

\(\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(B) - \phi(A)\)

This is analogous to FTC Part 2 for single-variable calculus.

Source: Wrede, R. C., & Spiegel, M. R. (2010).