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4.9 Applications of the Riemann Integral

After studying properties and some techniques to compute integrals, we now describe some applications.

Integrals are used in physics to compute several things, such as:

  • The work needed to move a particle
  • The position of a particle given its speed

This lecture covers two analytical applications:

  1. Computation of the area between graphs

  2. Length of a curve

Computing the Area Between Two Graphs

Let \(f, g: [a, b] \rightarrow \mathbb{R}\) be two integrable functions on the interval \([a, b]\).
Assume \(g(x)\geq f(x),\forall x\in[a,b]\). Define \(h(x) = g(x) - f(x)\) satisfying \(h(x)\geq0,\forall x\in[a,b]\), then the integral \(\displaystyle\int_{a}^{b}h(x)\,dx\)represents the area between the graphs of \(h\) and axis-\(x\).

In other words, \(\displaystyle\int_{a}^{b}h(x)\,dx = \int_{a}^{b}(g(x) - f(x))\,dx\) represents the area between the graphs \(g\) and the graphs of \(f\)image​

Examples

  1. \(f(x)=x,g(x)=x^3,x\in[0,1]\)​

    Find the intersection of \(f\) and \(g\). Find \(x\) with \(f(x)=g(x)\)​

    Then \(x=x^3\Rightarrow 0=x^3-x\), then \(x=0,x=1,x=-1\)

    By evaluating, we know \(f(x)\geq g(x)\)

    image​

    Why there isn't any other intersection?

    Consider \(h(x)=f(x)-g(x)\in[0,1]\), we know \(h(0)=0,h(1)=0\)​

    Then \(h'(x)=1-3x^2\), when \(h'(x)=0\), then \(x=\frac{1}{\sqrt3}\)​

    Thus \(h(x)\) increases in \([0,\frac{1}{\sqrt3})\)(\(h'(x)>0\)) and decreases in \([\frac{1}{\sqrt3},1]\)(\(h'(x)<0\))

    Then if there is another point such that \(f(x)=g(x)\Rightarrow h(x)=0\), then by Theorem (Rolle), there will be another point with \(h'(x)=0\) which is a contradiction

    Thus we have no more intersection

    Then \({{\displaystyle\int_0^1(f(x)-g(x))dx=\int_0^1(x-x^3)dx=\frac{x^{2}}{2}-\frac{x^{4}}{4}\Big|_0^1=\frac{1}{4}}}\)​

  2. \(f(x)=\cos(x),g(x)=\sin(x)\) in \([0,\pi]\)​

    Let's find the intersection

    \(f(x)=g(x)\Rightarrow\sin x-\cos x=0\Rightarrow\sqrt{2}\sin(x-\frac{\pi}{4})=0\Rightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=k\pi+\frac{\pi}{4}\)​

    And since \(x\in [0,\pi]\), then \(x=\frac{\pi}{4}\)

    Why there isn't any other intersections?

    Since \(f(x)\in[0,\frac{\pi}{4}]\) is decreasing and \(g(x)\in[0,\frac{\pi}{4}]\) is increasing and \(f(x)\leq g(x)\in[0,\frac{\pi}{4}]\), then the floor is \(\sin x\) and ceiling is \(\cos x\) in \([0,\frac{\pi}{4}]\)

    ......in \([\frac{\pi}{4},\frac{\pi}{2}]\)

    Thus \(Area =\) \(\displaystyle \int_{0}^{\frac{\pi}{4}}(\cos(x) - \sin(x)) \, dx + \displaystyle \int_{\frac{\pi}{4}}^{\pi}(\sin(x) - \cos(x)) \, dx\)

    \(= \sin(x) + \cos(x)\Big|_{0}^{\frac{\pi}{4}}+ -\cos(x) - \sin(x)\Big|_{\frac{\pi}{4}}^{\pi}\) \(=2\sqrt{2}\)

Length of a Curve

In this lecture, a curve is a smooth, or at least a differentiable function with continuous derivate (a \(C^1\) function or continuously differentiable).

\(C^1\) function is \(f\) is differentiable and \(f'\) is continuous

The idea is to use partitions and try to approximate the curve with a polygonal and compute the length of each segmentimage​

Let \(P = (x_0, x_1, \ldots, x_n)\) be a partition of \([a, b]\).

\(l(f, P) = \sum_{i=1}^{n} \text{distance from }(x_{i-1}, f(x_{i-1})) \text{ to }(x_{i}, f(x_{i}))\)

\(= \sum_{i=1}^n \sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}\)

\(= \sum_{i=1}^{n}\sqrt{1 + \underbrace{\left(\frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}\right)^{2}}_{f'(c_i) \text{ with }c_i \in [x_{i-1}, x_i]}\cdot (x_{i}- x_{i-1})}\)

Does these sums converge (in some sense) if we refine the partition?

If it does, we have the following definition.

Definition

Let \(f\) be a function on \([a, b]\). We say that the graph of \(f\) has a length in \([a, b]\) if there is \(l \in \mathbb{R}\) such that for all \(\epsilon > 0\), there exists \(\delta > 0\) such that for every partition \(P = (x_0, x_1, ..., x_n)\) with \(|x_i - x_{i-1}| < \delta\) we have \(|l(f, P) - l| < \epsilon\)

Theorem

If \(f\) is a continuously differentiable function on \([a, b]\), that is, \(f \in C^1([a, b])\), then the length of its graph in \([a, b]\) is \(\displaystyle \int_a^b \sqrt{1 + (f'(x))^2} \, dx\)

Example

Compute the length of the unit circle image​

\(S^1=\{(x,y)\in \R^2|x^2+y^2=1\}\)

Take \(f(x)=\sqrt{1-x^{2}}\), then \(f^{\prime}(x)=\frac{1}{2}\frac{1}{\sqrt{1-x^{2}}}\cdot(-2x)=-\frac{x}{\sqrt{1-x^{2}}}\)

\(f\) is differentiable and \(f'\) is continuous in \([0,1)\)

Then by theorem: \(l'=\displaystyle\int_{0}^{1}\sqrt{1+(f^{\prime}(x))^{2}}dx=\int_{0}^{1}\sqrt{1+(-\frac{x}{\sqrt{1-x^2}})^{2}} dx =\int_{0}^{1}\sqrt{\frac{1}{1-x^{2}}}dx\)

Then let \(x=\cos(t),1-x^2=\sin^2(x)\) and \(dx=-\sin(t)dt\)

Then we change the domain: \(1=\cos(0)\) and \(0=\cos(\frac{\pi}{2})\)

Then \(=\displaystyle \int_\frac{\pi}{2}^0\sqrt{\frac{1}{\sin^2(t)}}(-\sin(t))dt= -\int_\frac{\pi}{2}^0\frac{1}{\sin(t)}\sin(t)dt=- \int^\frac{\pi}{2}_01dt=\frac{\pi }{2}\)

Thus \(l=4l'=2\pi\)

Improper Integrals

Up to now, we have restricted our attention to bounded functions on a closed interval.

The reason is that an unbounded function may not be integrable and a function defined over an infinite interval brings up the question of convergence.

Now we extend the definition to include the cases where the function is unbounded or the interval is unbounded, or both.

Definition

We say that a function \(f\) is locally integrable on an interval \(I\) if \(f\) is integrable on every closed subinterval \([a, b] \subseteq I\).

For example,

  • \(f(x) = \sin(x)\) is locally integrable on \(I = (-\infty, \infty)\).
  • \(g(x) = \frac{1}{x}\) is locally integrable on \(I = (0, \infty)\).
  • \(h(x) = \frac{1}{x(x-1)}\) is locally integrable on the intervals \(I = (-\infty, 0)\), \(I = (0, 1)\) and \(I = (1, \infty)\).

Definition

Let \(f\) be a locally integrable function on \(I = [a, b)\). We define \(\displaystyle \int_{a}^{b}f(x) \, dx = \lim_{c \to b^-}\int_{a}^{c}f(x) \, dx = l\) in case the limit \(l\) exists.

Remark

The limit \(l\) above exists if the interval \([a, b]\) is finite (for example, if \(b \neq \infty\)) and the function is bounded (Exercise).

Proof

NTP: \(\forall\varepsilon>0,\exists\delta:c\in\left(b-\delta,b\right):\left|{{\displaystyle\int_{a}^{c}f(x)\,dx-\int_{a}^{b}f(x)\,dx}} \right|<\varepsilon\)

Since \(f(x)\) is bounded, then \(m\left(b-c\right)\leq{{\displaystyle\int_{c}^{b}f(x)\,dx\leq M(b-c)}}\)

Then \(\left|{{{\displaystyle\int_{c}^{b}f(x)\,dx}}}\right|\leq k\left(b-c\right)=\min\left (\left|m\left(b-c\right)\right|,\left|M\left(b-c\right)\right|\right)\)

\(\left|{{{{{{{\displaystyle\int_{a}^{c}f(x)\,dx-\int_{a}^{b}f(x)\,dx}}}}}}}\right |=\left|{{{{{\displaystyle\int_{c}^{b}f(x)\,dx}}}}}\right|\leq k\left(b-c\right)\)

And we know \(b-\delta<c<b\), then \(0<b-c<\delta\)​

Then \(\left|{{{{{{{{\displaystyle\int_{a}^{c}f(x)\,dx-\int_{a}^{b}f(x)\,dx}}}}}}}}\right |<k\delta\)

Then if \(k=0\), it's clearly less than \(\varepsilon\). If \(k\neq 0\), then take \(\delta <\frac{\varepsilon}{k}\)

However, the limit may also exists in cases where \(b = \infty\) or \(b = \infty\) and the function is unbounded.

In such a case, the limit above assigns a value to an integral which cannot be properly defined with our definition.

For this reason, we call it improper integral, and we say that \(\displaystyle \int_a^b f(x) \, dx\) converges to the value \(l\)

We say that \(f\) is integrable on \([a, b]\) if the limit \(l\) exists.

If it does not exists, we say that \(\displaystyle \int_{a}^{b}f(x) \, dx\) diverges and \(f\) is non-integrable on \([a, b]\).

Similarly as above, we have the following definition.

Definition

Let \(f\) be a locally integrable function on \(I = (a, b]\). We define \(\displaystyle \int_a^b f(x) \, dx = \lim_{c \to a^+} \int_c^b f(x) \, dx = l\), in case the limit \(l\) exists.

Example

  1. \(\displaystyle\int_{1}^{\infty}\frac{1}{x^{2}}dx = \lim_{c \to \infty}\int_{1}^{c} \frac{1}{x^{2}}dx\) where \(I=[1,\infty)\)

    We know \(f(x)=\frac{1}{x^2}\) is integrable in \([1,c]\)​

    \(\displaystyle\int_{1}^{c} \frac{1}{x^2} dx = \displaystyle\int_{1}^{c} x^{-2} dx = -x^{-1}\Big|_1^c = -\frac{1}{c} - (-1) = -\frac{1}{c} + 1\)

    Then \(\displaystyle\lim_{c \to \infty} \left(-\frac{1}{c} + 1\right) = 1\). Thus\(\displaystyle \int_{1}^{\infty}\frac{1}{x^{2}}dx\) converges to \(1\)

  2. \(\displaystyle\int_{0}^{1} \frac{1}{\sqrt{x}} dx\) where \(I = (0, 1]\)

    Then consider \(\displaystyle\lim_{c \to 0^+} \int_{c}^{1} \frac{1}{\sqrt{x}} dx\), we have \(\displaystyle\int_{c}^{1} \frac{1}{\sqrt{x}} dx = \displaystyle\int_{c}^{1} x^{-1/2} dx = \frac{x^{1/2}}{1/2}\Big|_c^1 = 2\sqrt{x}\Big|_c^1 = 2 - 2\sqrt{c}\)

    Then \(\displaystyle\lim_{c \to 0^+} 2 - 2\sqrt{c} = 2\), thus \(\displaystyle\int_{0}^{1}\frac{1}{\sqrt{x}}dx\) converges to \(2\)​

  3. \(\displaystyle\int_{-1}^{1} \frac{1}{x} dx\) where \(I = (-1, 0) \cup (0, 1)\)

    Then we need to separate \(\displaystyle\int_{-1}^{0} \frac{1}{x} dx\), \(\displaystyle\int_{0}^{1} \frac{1}{x} dx\)