--- title: "Integration" book: "High School Mathematics" subject: math language: en chapter: 25 exercises: 9 source: https://one-course.com/books/math/2/en/chapter/25-integration --- # Chapter 25 — Integration Integration answers two questions at once: what is the area under a curve, and how can one recover a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) from its rate of change? The fundamental theorem of calculus states that these are the same question — the deepest and most useful discovery of seventeenth-century mathematics. ## 25.1 The integral of a continuous function **Definition 25.1 (Integral of a positive function).** Let $f$ be [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) and positive on $\intcc{a}{b}$. The *integral* $$ \int_a^b f(x)\,\dd x $$ is the area, in units of area, of the region bounded by the curve of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$. ![](https://one-course.com/images/onecourse/chapters/math-2/g12-integ/fig-be957e4dcae1.svg) For a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) of arbitrary sign, areas below the $x$-axis are counted negatively; and one sets $\int_b^a f(x)\,\dd x = -\int_a^b f(x)\,\dd x$. ![Signed areas: ∈t_02π x\, x = 0, the region below the axis (red) cancelling the region above (blue).](https://one-course.com/images/onecourse/chapters/math-2/g12-integ/fig-49f37058cc51.svg) *Signed areas: $\int_0^{2\pi} \sin x\,\dd x = 0$, the region below the axis (red) cancelling the region above (blue).* **Proposition 25.2 (Properties of the integral).** Let $f, g$ be [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) containing $a, b, c$, and $\lambda \in \R$. 1. *Linearity:* $\displaystyle\int_a^b (f + \lambda g) = \int_a^b f + \lambda \int_a^b g$ . 2. *Chasles relation:* $\displaystyle\int_a^c f = \int_a^b f + \int_b^c f$ . 3. *Positivity:* if $a \leq b$ and $f \geq 0$ on $\intcc{a}{b}$ , then $\displaystyle\int_a^b f \geq 0$ ; if $f \leq g$ , then $\displaystyle\int_a^b f \leq \int_a^b g$ . **Proof.** Chasles and positivity are immediate from the area interpretation (areas add up when regions are juxtaposed; a region of positive height has positive area). The comparison follows by applying positivity to $g - f$. Linearity is intuitively clear for the sum of positive [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) (stack the areas) and is proved rigorously at university; *it is admitted here*. ∎ ## 25.2 The fundamental theorem of calculus **Definition 25.3 (Primitive).** A *primitive* (or antiderivative) of $f$ on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ is a [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $F$ on $I$ such that $F' = f$. **Proposition 25.4.** If $F$ is a [primitive](#def-g12-integ-primitive) of $f$ on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$, the [primitives](#def-g12-integ-primitive) of $f$ on $I$ are exactly the [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $F + c$, $c \in \R$. Given $x_0 \in I$ and $y_0 \in \R$, there is a unique [primitive](#def-g12-integ-primitive) with $F(x_0) = y_0$. **Proof.** If $G' = F' = f$, then $(G - F)' = 0$ on the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$, so $G - F$ is constant.[1](#fn-1) Conversely every $F + c$ is a [primitive](#def-g12-integ-primitive). The condition $F(x_0) = y_0$ pins down the constant. ∎ **Theorem 25.5 (Fundamental theorem of calculus).** Let $f$ be [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ and $a \in I$. The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $$ F \colon x \longmapsto \int_a^x f(t)\,\dd t $$ is the [primitive](#def-g12-integ-primitive) of $f$ on $I$ vanishing at $a$. Consequently, for any [primitive](#def-g12-integ-primitive) $G$ of $f$ and $a, b \in I$: $$ \int_a^b f(t)\,\dd t = \bigl[G(t)\bigr]_a^b = G(b) - G(a). $$ ![The idea of the proof: the increment F(x+h) - F(x) is the area of the narrow strip (orange), squeezed between the rectangles of heights f(x) and f(x+h) over a base of length h.](https://one-course.com/images/onecourse/chapters/math-2/g12-integ/fig-fc94b388b117.svg) *The idea of the proof: the increment $F(x+h) - F(x)$ is the area of the narrow strip (orange), squeezed between the rectangles of heights $f(x)$ and $f(x+h)$ over a base of length $h$.* **Proof when $f$ is increasing.** Fix $x \in I$ and $h > 0$ with $x + h \in I$. By Chasles, $$ F(x+h) - F(x) = \int_x^{x+h} f(t)\,\dd t . $$ Since $f$ is [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), $f(x) \leq f(t) \leq f(x+h)$ for $t \in \intcc{x}{x+h}$, and by comparison of [integrals](#def-g12-integ-area) (the [integral](#def-g12-integ-area) of a constant $c$ over an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of length $h$ is $ch$): $$ h\,f(x) \leq F(x+h) - F(x) \leq h\,f(x+h), $$ so $$ f(x) \leq \frac{F(x+h) - F(x)}{h} \leq f(x+h). $$ As $h \to 0^+$, $f(x+h) \to f(x)$ by [continuity](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity), and the squeeze theorem gives that the difference quotient tends to $f(x)$; the case $h < 0$ is symmetric. Hence $F' = f$, and $F(a) = 0$. The general ([non-monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic)) [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) case is proved at university. Finally, if $G$ is any [primitive](#def-g12-integ-primitive), $G = F + c$ ([Proposition 25.4](#prop-g12-integ-primunique)), so $G(b) - G(a) = F(b) - F(a) = \int_a^b f$. ∎ **Example 25.6.** $\displaystyle\int_0^1 x^2\,\dd x = \left[\frac{x^3}{3}\right]_0^1 = \frac13$: the area under the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) is one third of the unit square, as Archimedes knew. The table of [primitives](#def-g12-integ-primitive) is read from the table of [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) ($u$ denotes a [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function), $c$ an arbitrary constant): | $f(x)$ | [primitive](#def-g12-integ-primitive) | $f$ | [primitive](#def-g12-integ-primitive) | | --- | --- | --- | --- | | $x^n \ (n \neq -1)$ | $\dfrac{x^{n+1}}{n+1} + c$ | $u'u^n \ (n \neq -1)$ | $\dfrac{u^{n+1}}{n+1} + c$ | | $\dfrac1x \ (x > 0)$ | $\ln x + c$ | $\dfrac{u'}{u} \ (u > 0)$ | $\ln u + c$ | | $\eu^x$ | $\eu^x + c$ | $u'\eu^u$ | $\eu^u + c$ | | $\cos x$ | $\sin x + c$ | $\dfrac{u'}{\sqrt u} \ (u>0)$ | $2\sqrt u + c$ | | $\sin x$ | $-\cos x + c$ | | | **Method 25.7 (Recognizing the form u′×(something in u)u' \times (\text{something in } u)u′×(something in u)).** To integrate a product, look for a factor which is the [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) of an inner [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $u$, up to a multiplicative constant. For instance in $\int_0^1 x\,\eu^{x^2}\dd x$, the factor $x$ is $\frac12 (x^2)'$: $$ \int_0^1 x\,\eu^{x^2}\dd x = \frac12\left[\eu^{x^2}\right]_0^1 = \frac{\eu - 1}{2}. $$ **Theorem 25.8 (Integration by parts).** Let $u, v$ be [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) on $\intcc{a}{b}$ with [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative). Then $$ \int_a^b u'(t)\,v(t)\,\dd t = \bigl[u(t)\,v(t)\bigr]_a^b - \int_a^b u(t)\,v'(t)\,\dd t . $$ **Proof.** The product rule gives $(uv)' = u'v + uv'$; integrating both sides over $\intcc{a}{b}$ and using the fundamental theorem for the left-hand side yields $\bigl[uv\bigr]_a^b = \int_a^b u'v + \int_a^b uv'$. ∎ **Example 25.9.** $\displaystyle\int_0^1 t\,\eu^{t}\,\dd t$: take $u' = \eu^t$, $v = t$, so $u = \eu^t$, $v' = 1$: $$ \int_0^1 t\,\eu^t\,\dd t = \bigl[t\,\eu^t\bigr]_0^1 - \int_0^1 \eu^t\,\dd t = \eu - (\eu - 1) = 1 . $$ ## 25.3 Applications **Definition 25.10 (Mean value).** The *mean value* of a [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $f$ on $\intcc{a}{b}$ ($a 0$. ∎ **Method 25.12 (Area between two curves).** If $f \geq g$ on $\intcc{a}{b}$, the area between the two curves is $\int_a^b \bigl(f(x) - g(x)\bigr)\dd x$. If the curves cross, split the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) at the crossing points and integrate $\abs{f - g}$ piece by piece. ![The area between the line y = x + 1 (red) and the parabola y = x2 - 1 (blue) is ∈t_-12 ((x+1) - (x2-1) ) x = 9/2.](https://one-course.com/images/onecourse/chapters/math-2/g12-integ/fig-d9b9464c2389.svg) *The area between the line $y = x + 1$ (red) and the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) $y = x^2 - 1$ (blue) is $\int_{-1}^{2} \bigl((x+1) - (x^2-1)\bigr)\dd x = \frac92$.* ## 25.4 Exercises **Exercise 25.1 ★.** Compute $$ \int_1^2 \left(3x^2 - \frac{1}{x^2}\right)\dd x, \qquad \int_0^{\pi/2} \cos t \,\dd t, \qquad \int_0^{1} \frac{\dd t}{2t+1} . $$ **Solution of Exercise 25.1.** $$ \int_1^2 \left(3x^2 - \frac{1}{x^2}\right)\dd x = \left[x^3 + \frac1x\right]_1^2 = \left(8 + \tfrac12\right) - 2 = \frac{13}{2}. $$ $$ \int_0^{\pi/2} \cos t\,\dd t = \bigl[\sin t\bigr]_0^{\pi/2} = 1 . $$ With $u = 2t+1$, $\frac{1}{2t+1} = \frac12\,\frac{u'}{u}$: $$ \int_0^1 \frac{\dd t}{2t+1} = \frac12\bigl[\ln(2t+1)\bigr]_0^1 = \frac{\ln 3}{2}. $$ **Exercise 25.2 ★.** Find the [primitive](#def-g12-integ-primitive) $F$ of $f(x) = x\eu^{x^2}$ on $\R$ such that $F(0) = 1$. **Solution of Exercise 25.2.** $x\,\eu^{x^2} = \frac12\,(x^2)'\,\eu^{x^2}$ has [primitives](#def-g12-integ-primitive) $\frac12\eu^{x^2} + c$. The condition $F(0) = 1$ gives $\frac12 + c = 1$, so $$ F(x) = \frac{\eu^{x^2} + 1}{2}. $$ **Exercise 25.3 ★.** Compute the [mean value](#def-g12-integ-mean) of $f(t) = \sin t$ on $\intcc{0}{\pi}$, and interpret the result on a [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph). **Solution of Exercise 25.3.** $$ \mu = \frac{1}{\pi}\int_0^\pi \sin t\,\dd t = \frac{1}{\pi}\bigl[-\cos t\bigr]_0^\pi = \frac{2}{\pi} \approx 0.64 . $$ Graphically, the rectangle of base $\intcc{0}{\pi}$ and height $\frac{2}{\pi}$ has the same area as the arch of the [sine](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#def-g12-trigo-cossin) curve. **Exercise 25.4 ★★.** Using [integration by parts](#thm-g12-integ-ibp), compute $$ \int_1^{\eu} \ln t\,\dd t \qquad\text{and}\qquad \int_0^{\pi} t \sin t\,\dd t . $$ **Solution of Exercise 25.4.** Take $u' = 1$, $v = \ln t$, so $u = t$, $v' = \frac1t$: $$ \int_1^{\eu} \ln t\,\dd t = \bigl[t\ln t\bigr]_1^{\eu} - \int_1^{\eu} 1\,\dd t = \eu - (\eu - 1) = 1 . $$ Take $u' = \sin t$, $v = t$, so $u = -\cos t$, $v' = 1$: $$ \int_0^\pi t\sin t\,\dd t = \bigl[-t\cos t\bigr]_0^\pi + \int_0^\pi \cos t\,\dd t = \pi + 0 = \pi . $$ **Exercise 25.5 ★★.** Compute the area of the region between the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) $y = x^2$ and the line $y = x + 2$. **Solution of Exercise 25.5.** The curves intersect where $x^2 = x + 2$, *i.e.* $x \in \{-1, 2\}$, and on $\intcc{-1}{2}$ the line is above the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola). The area is $$ \int_{-1}^{2} \bigl(x + 2 - x^2\bigr)\dd x = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}. $$ **Exercise 25.6 ★★.** Let $I = \displaystyle\int_0^1 \frac{\dd t}{1 + t}$. 1. Compute $I$ . 2. For $n \in \N$ , let $I_n = \displaystyle\int_0^1 \frac{t^n}{1+t}\dd t$ . Show that $I_n + I_{n+1} = \dfrac{1}{n+1}$ , and that $0 \leq I_n \leq \dfrac{1}{n+1}$ . 3. Deduce that $1 - \frac12 + \frac13 - \dots + \frac{(-1)^{n-1}}{n} \xrightarrow[n \to +\infty]{} \ln 2$ . **Solution of Exercise 25.6.** *1.* $I = \bigl[\ln(1+t)\bigr]_0^1 = \ln 2$. *2.* $$ I_n + I_{n+1} = \int_0^1 \frac{t^n(1 + t)}{1+t}\,\dd t = \int_0^1 t^n \,\dd t = \frac{1}{n+1}. $$ On $\intcc{0}{1}$, $0 \leq \dfrac{t^n}{1+t} \leq t^n$, so by comparison $0 \leq I_n \leq \frac{1}{n+1}$. *3.* From the recurrence, by a telescoping induction, $$ I_n = (-1)^n\left(I_0 - \left(1 - \frac12 + \dots + \frac{(-1)^{n-1}}{n}\right)\right), $$ *i.e.* $\displaystyle\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} = \ln 2 - (-1)^n I_n$ (recall $I_0 = \ln 2$). Since $\abs{I_n} \leq \frac{1}{n+1} \to 0$, the alternating harmonic sum tends to $\ln 2$. **Exercise 25.7 ★★.** The speed of a train (in m/s) during the first $100$ seconds after departure is modeled by $v(t) = 30\bigl(1 - \eu^{-t/50}\bigr)$. Compute the distance traveled during these $100$ seconds, and the train’s average speed over the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval). **Solution of Exercise 25.7.** Distance: $$ d = \int_0^{100} 30\left(1 - \eu^{-t/50}\right)\dd t = 30\left[t + 50\,\eu^{-t/50}\right]_0^{100} = 30\left(100 + 50\eu^{-2} - 50\right) = 1500 + 1500\,\eu^{-2}. $$ Numerically $d \approx 1500 + 203 = 1703$ m. Average speed: $\frac{d}{100} = 15\bigl(1 + \eu^{-2}\bigr) \approx 17.0$ m/s. **Exercise 25.8 ★★★.** For $n \geq 1$, let $S_n = \dfrac1n \displaystyle\sum_{k=1}^{n} \frac{1}{1 + k/n}$. 1. Interpret $S_n$ as an area of rectangles approximating a region under the curve of $t \mapsto \frac{1}{1+t}$ on $\intcc{0}{1}$ . 2. Using the monotonicity of $t \mapsto \frac{1}{1+t}$, show that $$S_n \leq \int_0^1 \frac{\dd t}{1+t} \leq S_n + \frac1n\left(1 - \frac12\right),$$ and deduce $\lim\limits_{n\to+\infty} S_n = \ln 2$. **Solution of Exercise 25.8.** *1.* $S_n$ is the total area of $n$ rectangles of width $\frac1n$ and heights $f\left(\frac kn\right)$, $k = 1, \dots, n$, where $f(t) = \frac{1}{1+t}$: it is the “right endpoint” rectangle [approximation](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-approx) of the area under $f$ on $\intcc{0}{1}$. *2.* $f$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations), so on each [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intcc{\frac{k-1}{n}}{\frac kn}$, $f\left(\frac kn\right) \leq f(t) \leq f\left(\frac{k-1}{n}\right)$; integrating and summing over $k$: $$ S_n \leq \int_0^1 f(t)\,\dd t \leq S_n + \frac1n\bigl(f(0) - f(1)\bigr) = S_n + \frac{1}{2n}. $$ (The middle term of the exercise’s display, with $f(0) - f(1) = 1 - \frac12$, is exactly this bound.) Hence $0 \leq \ln 2 - S_n \leq \frac{1}{2n} \to 0$, so $S_n \to \ln 2$. **Exercise 25.9 ★★★.** *(Wallis [integrals](#def-g12-integ-area).)* For $n \in \N$, let $W_n = \displaystyle\int_0^{\pi/2} \sin^n t\,\dd t$. 1. Compute $W_0$ and $W_1$ . 2. Writing $\sin^{n+2}t = \sin t \cdot \sin^{n+1} t$ and integrating by parts, show that $W_{n+2} = \dfrac{n+1}{n+2}\,W_n$ . 3. Deduce $W_2$ , $W_3$ , $W_4$ and show that $(W_n)$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) and positive. **Solution of Exercise 25.9.** *1.* $W_0 = \int_0^{\pi/2} \dd t = \frac{\pi}{2}$ and $W_1 = \bigl[-\cos t\bigr]_0^{\pi/2} = 1$. *2.* Integrate by parts with $u' = \sin t$, $v = \sin^{n+1} t$, so $u = -\cos t$, $v' = (n+1)\sin^n t \cos t$: $$ W_{n+2} = \bigl[-\cos t \sin^{n+1} t\bigr]_0^{\pi/2} + (n+1)\int_0^{\pi/2} \cos^2 t\,\sin^n t\,\dd t = (n+1)\int_0^{\pi/2} (1 - \sin^2 t)\sin^n t\,\dd t, $$ the bracket vanishing at both ends. Hence $W_{n+2} = (n+1)(W_n - W_{n+2})$, *i.e.* $W_{n+2} = \frac{n+1}{n+2} W_n$. *3.* $W_2 = \frac12 W_0 = \frac{\pi}{4}$, $W_3 = \frac23 W_1 = \frac23$, $W_4 = \frac34 W_2 = \frac{3\pi}{16}$. On $\intoo{0}{\frac\pi2}$, $0 < \sin t < 1$, so $\sin^{n+1} t \leq \sin^n t$ with strict inequality on the interior; integrating, $0 < W_{n+1} \leq W_n$ (in fact $<$): $(W_n)$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) and positive. ## 25.5 Problem: Archimedes against the machine **Problem 25.1.** Weekend problem — the area under the parabola, computed three ways across twenty-two centuries, and the logarithm’s secret identity as an area Around 240 BC, Archimedes computed the exact area of a parabolic segment — without [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), without limits, without algebra. Nineteen centuries later, Riemann’s rectangles redid it by brute squeeze; and the fundamental theorem of calculus ([Theorem 25.5](#thm-g12-integ-ftc)) now does it in one line. This problem plays all three matches, then uses the same machine to reveal what the [logarithm](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln) really is: an area with a scaling symmetry. **Part I — Fluency.** 1. Compute $\displaystyle\int_0^1 (3x^2 - 2x + 1)\,\dd x$ , $\displaystyle\int_1^{\eu} \frac{\dd x}{x}$ , and $\displaystyle\int_0^{\pi/2} \cos x\,\dd x$ . 2. Spot the $u'u$ -forms ( [Method 25.7](#met-g12-integ-uprime) ): $\displaystyle\int_0^1 x\,\eu^{x^2}\dd x$ and $\displaystyle\int_0^1 \frac{2x}{x^2 + 1}\,\dd x$ . 3. By parts ( [Theorem 25.8](#thm-g12-integ-ibp) ): $\displaystyle\int_0^1 x\,\eu^x \dd x$ and $\displaystyle\int_1^{\eu} \ln x\,\dd x$ . 4. Compute the [mean value](#def-g12-integ-mean) ( [Definition 25.10](#def-g12-integ-mean) ) of $\sin$ over $\intcc{0}{\pi}$ — and note it is *not* $\frac12$ . 5. Compute the area between the line $y = x$ and the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) $y = x^2$ over $\intcc{0}{1}$ ( [Method 25.12](#met-g12-integ-areabetween) ). **Part II — The [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola), three ways.** 6. Riemann’s setup for the area under $y = x^2$ on $\intcc{0}{1}$ : write the lower sum $L_n$ and the upper sum $U_n$ over $n$ equal rectangles (as in [Exercise 25.8](#exo-g12-integ-8) ). 7. Prove by induction ([Problem 20.1](https://one-course.com/books/math/2/en/chapter/20-sequences#pb-g12-seq-1)’s machine) the sum-of-squares formula $$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} .$$ 8. Deduce closed forms for $U_n$ and $L_n$ , compute their common limit, and conclude: the area is $\frac13$ . 9. Now the machine: compute $\int_0^1 x^2\,\dd x$ by the fundamental theorem, in one line. Compare the efforts. 10. Archimedes stated it otherwise: *a parabolic segment is $\frac43$ of its inscribed triangle* . For the segment cut from $y = x^2$ by the chord joining $(-1, 1)$ to $(1, 1)$ : compute the segment’s area with an [integral](#def-g12-integ-area) , the inscribed triangle’s area (apex at the vertex $(0, 0)$ ), and verify the master’s ratio. 11. Archimedes’ own method: fill the segment with the big triangle $T$ , then two triangles totalling $\frac T4$ , then four totalling $\frac{T}{16}$ , and so on. Sum the geometric series and recover $\frac43 T$ — the endlessly bitten chocolate bar of the Middle School volume, eaten by a Greek geometer. 12. One sentence each: exhaustion (Archimedes), squeeze (Riemann), anti-differentiation (Newton–Leibniz) — what does each need, and what does each give? **Part III — The [logarithm](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln) is an area.** For $x > 0$, set $A(x) = \displaystyle\int_1^x \frac{\dd t}{t}$. 13. Give $A(1)$ and $A'(x)$ ( [Theorem 25.5](#thm-g12-integ-ftc) ), and conclude that $A$ is exactly the [natural logarithm](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln) of [Definition 23.5](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln) . 14. The scaling miracle: fix $a > 0$ and study $g(x) = A(ax) - A(x)$. Compute $g'$ (chain rule), deduce that $g$ is constant, evaluate the constant — and conclude the functional [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $$\ln(ab) = \ln a + \ln b ,$$ proved by pure calculus: the area from $1$ to $ab$ splits into scaled copies. 15. Deduce from question 14: $\ln(a^n) = n\ln a$ and $\ln\frac1a = -\ln a$ . 16. [Exercise 25.8](#exo-g12-integ-8) reads $\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}$ as rectangles under $\frac{1}{1 + t}$ : compute this sum for $n = 10$ (three decimals) and compare with $\ln 2$ . Which harmonic-flavored sums, divergent term by term, here converge to an area? **Part IV — Accumulation.** 17. A car accelerates with speed $v(t) = 3t^2$ m/s for $t \in \intcc{0}{10}$ . Compute the distance covered, the [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) speed, and the instant at which the instantaneous speed equals the [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) speed. 18. A $4$ -meter rod has linear density $\rho(x) = 2 + x$ kg/m. Compute its total mass and its center of mass $\dfrac{\int_0^4 x\,\rho(x)\,\dd x}{\int_0^4 \rho(x)\,\dd x}$ — the balance point of the Middle School volume’s cardboard triangle, finally computed with weights that vary. 19. Wallis’s [integrals](#def-g12-integ-area) ( [Exercise 25.9](#exo-g12-integ-9) ): compute $W_0$ , $W_1$ , and $W_2 = \int_0^{\pi/2} \sin^2 t\,\dd t$ using the linearization of [Problem 24.1](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#pb-g12-trigo-1) . (Their infinite ladder climbs, in the university volumes, to a product formula for $\pi$ and to the normalization of the bell curve.) 20. Finale — integration’s three faces: an *area* by definition, an *accumulation* in the world (distance, mass), an *[anti-derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative)* by the fundamental theorem; and its history in three names. Close with the chapter’s pearl: which everyday button on the calculator is secretly the area under $\frac1t$ — and which identity did that area prove? **Solution of Problem 25.1.** **1.** $\left[x^3 - x^2 + x\right]_0^1 = 1$; $\left[\ln x\right]_1^{\eu} = 1$; $\left[\sin x\right]_0^{\pi/2} = 1$. **2.** $\left[\frac12 \eu^{x^2}\right]_0^1 = \frac{\eu - 1}{2}$; $\left[\ln(x^2 + 1)\right]_0^1 = \ln 2$. **3.** $\int_0^1 x\eu^x = \left[x\eu^x\right]_0^1 - \int_0^1 \eu^x = \eu - (\eu - 1) = 1$. And $\int_1^{\eu} \ln x = \left[x\ln x\right]_1^{\eu} - \int_1^{\eu} 1 = \eu - (\eu - 1) = 1$. **4.** $\frac{1}{\pi}\int_0^\pi \sin x\,\dd x = \frac{2}{\pi} \approx 0.64$: the average height of the [sine](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#def-g12-trigo-cossin) arch is $\frac2\pi$, noticeably more than one half — the arch is fat near its top. **5.** $\int_0^1 (x - x^2)\,\dd x = \frac12 - \frac13 = \frac16$. **6.** $L_n = \frac1n \sum_{k=0}^{n-1} \left(\frac kn\right)^2$ and $U_n = \frac1n \sum_{k=1}^{n} \left(\frac kn\right)^2$: rectangles of width $\frac1n$ with heights read at the left (below the curve) or right (above) endpoints. **7.** True for $n = 1$ ($1 = \frac{1 \cdot 2 \cdot 3}{6}$). If it holds for $n$, then adding $(n+1)^2$: $$ \frac{n(n+1)(2n+1)}{6} + (n+1)^2 = \frac{(n+1)\left(2n^2 + n + 6n + 6\right)}{6} = \frac{(n+1)(n+2)(2n+3)}{6}, $$ the formula at $n + 1$: heredity, done. **8.** $U_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(1 + \frac1n)(2 + \frac1n)}{6} \to \frac26 = \frac13$, and $L_n = U_n - \frac1n \to \frac13$: squeezed between its staircases, the area is exactly $\frac13$. **9.** $\int_0^1 x^2\,\dd x = \left[\frac{x^3}{3}\right]_0^1 = \frac13$: one line. The fundamental theorem converts an infinite squeeze into a single [anti-derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) evaluation — that is why it is called fundamental. **10.** Segment: $\int_{-1}^{1}(1 - x^2)\,\dd x = 2 - \frac23 = \frac43$. Triangle with base the chord (length $2$, at height $1$) and apex $(0,0)$: area $\frac12 \times 2 \times 1 = 1$. Ratio: $\frac43$ — exactly Archimedes’ theorem, verified by the machine he lacked. **11.** $T\left(1 + \frac14 + \frac{1}{16} + \dots\right) = T \cdot \frac{1}{1 - \frac14} = \frac43 T$ (geometric series of ratio $\frac14$): with $T = 1$, the segment’s area $\frac43$ again. Archimedes summed the series by a purely geometric argument — the quarter-bites of the chocolate-bar kind, two millennia before limits were words. **12.** Exhaustion needs a clever geometric self-similarity and gives exact areas case by case; Riemann needs only monotonicity and gives a universal definition at the price of computing sums; the fundamental theorem needs the whole edifice of [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) and repays it by making areas one-line computations. **13.** $A(1) = 0$ and $A'(x) = \frac1x$ (fundamental theorem): $A$ is the [primitive](#def-g12-integ-primitive) of $\frac1x$ on $\intoo{0}{+\infty}$ vanishing at $1$ — which is precisely the [natural logarithm](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln). **14.** $g'(x) = \frac{a}{ax} - \frac1x = 0$: $g$ is constant; at $x = 1$: $g(1) = A(a) - A(1) = A(a)$. Hence $A(ax) = A(a) + A(x)$ for all $x$, and with $x = b$: $\ln(ab) = \ln a + \ln b$. The [hyperbola](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-reference)’s area does not care about horizontal scale: stretching $t$ by $a$ squashes $\frac1t$ by $a$, and the area survives — addition of [logarithms](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#def-g12-exp-ln) is a scaling symmetry. **15.** Iterating: $\ln(a^n) = n\ln a$ (induction on $n$). And $0 = \ln 1 = \ln\left(a \cdot \frac1a\right) = \ln a + \ln\frac1a$: $\ln\frac1a = -\ln a$. **16.** $\frac{1}{11} + \frac{1}{12} + \dots + \frac{1}{20} \approx 0.669$, against $\ln 2 \approx 0.693$: ten rectangles under $\frac{1}{1+t}$, already close. The harmonic series diverges, but its *slices* from $n$ to $2n$ settle on $\ln 2$: divergence at a crawl, measured by an area. **17.** Distance: $\int_0^{10} 3t^2\,\dd t = 1000$ m. [Mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) speed: $100$ m/s. Instant with $v = 100$: $3t^2 = 100$: $t = \sqrt{\frac{100}{3}} \approx 5.77$ s. **18.** Mass: $\int_0^4 (2 + x)\,\dd x = 8 + 8 = 16$ kg. Moment: $\int_0^4 (2x + x^2)\,\dd x = 16 + \frac{64}{3} = \frac{112}{3}$. Center of mass: $\frac{112/3}{16} = \frac73 \approx 2.33$ m — pushed beyond the middle by the heavier right end, as intuition demands. **19.** $W_0 = \frac\pi2$; $W_1 = \left[-\cos t\right]_0^{\pi/2} = 1$; and $W_2 = \int_0^{\pi/2} \frac{1 - \cos 2t}{2}\,\dd t = \frac\pi4$. **20.** Area: the definition, squeezed by staircases. Accumulation: distances from speeds, masses from densities — the [integral](#def-g12-integ-area) as a running total. [Anti-derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative): the fundamental theorem, Newton’s and Leibniz’s shortcut through Archimedes’ patience. The pearl: the $\ln$ button is the area under $\frac1t$, and that area’s scaling symmetry *is* the identity $\ln(ab) = \ln a + \ln b$. 1. That a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) with zero [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) is constant follows from [Theorem 22.7](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#thm-g12-deriv-variations) : it is both [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) and [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) . [↩](#fnref-1)