# Integration by parts.

Success is not final, failure is not fatal: It is the courage to continue that counts, Winston Churchill.

# Integration by parts

It is a technique used in calculus to evaluate integrals of the form $\int uv’dx$. Recall that (uv)’ = u’v + uv’ ⇒ uv’ = (uv)’ -u’v ⇒ $\int uv’dx = uv -\int u’vdx$, e.g., $\int lnx dx =$[u = lnx, u’ = 1/x, v = x, v’ = 1] xlnx -$\int \frac{1}{x}xdx = xlnx -x + C.$ 🚀

• $\int (lnx)^2 dx =$[u = (lnx)2, u’=2(lnx)(1/x), v = x, v’ = 1] x(lnx)2 -$\int 2(lnx)\frac{1}{x}xdx = x(lnx)^2 -\int 2(lnx)dx$ =[By the previous exercise, 🚀] x(lnx)2 -2(xlnx -x) + C. 🚂

• In the general case, $\int (lnx)^n dx =$[u = (lnx)n, u’ = n(lnx)n-1(1/x), v = x, v’ = 2] x(lnx)n -n$\int (lnx)^{n-1}\frac{1}{x}x dx = x(lnx)^n -n\int (lnx)^{n-1}dx$.

If we define Fn(x) = $\int (lnx)^n dx$, Fn(x) = x(lnx)n -nFn-1(x). Hence, F0(x) = $\int (lnx)^0 dx = x$, F1(x) = x(lnx) -F0(x) = xlnx -x + C🚀, F2(x) = x(lnx)2 -2F1(x) = x(lnx)2 -2(xlnx -x) + C 🚂

$\int x^ne^xdx$ =[u = xn, u’ = nxn-1, v’ = ex ⇒ v = ex] xnex -$\int nx^{n-1}e^xdx$, so Fn(x) = $\int x^ne^xdx = x^ne^x -nF_{n-1}(x)$.

F0(x) = $\int e^xdx = e^x + C.$ F1(x) = $\int xe^xdx$ = xex -ex + C, and so on…

• Find the volume of an exponential wineglass, that is, y = ex rotated around the y-axis between y = 1 and y = e.

The answer is -horizontal slicing- $\int_{1}^{e} πx^2dy$ =[y = ex, x = lny] π$\int_{1}^{e} (lny)^2dy = πF_2(y)\bigg|_{1}^{e}$ =

Considering the previous exercise, = $π(y(lny)^2-2(ylny -y)) \bigg|_{1}^{e}$ = π(e-2).

Analogously, the answer is -vertical slicing- $\int_{0}^{1} (e-y)2πxdx = \int_{0}^{1} (e-e^x)2πxdx = \int_{0}^{1}2πexdx -2π\int_{0}^{1} xe^xdx = 2πe\frac{x^2}{2} -2πxe^x +2πe^x\bigg|_{0}^{1} = 2πe\frac{1}{2}-2πe+2πe-2π = πe -2π = π(e-2).$

• $\int xtan^{-1}(x)dx$ =[v’=x, u = tan-1x, u’ = c, v = x²/2] = $\frac{x^2}{2}tan^{-1}(x) - \int \frac{x^2}{2}\frac{1}{1+x^2}dx = \frac{x^2}{2}tan^{-1}(x) - \frac{1}{2}\int \frac{x^2}{1+x^2}dx$ = [1+x² = x²+1-11+x² = 1-11+x² ] = $\frac{x^2}{2}tan^{-1}(x)-\frac{1}{2}x+\frac{1}{2}tan^{-1}x + C.$

• Prove that $\int_{0}^{∞} x^ne^{-x}dx = n!$ for all n ≥ 0.

Case base, n = 0, $\int_{0}^{∞} x^ne^{-x}dx = \int_{0}^{∞} e^{-x}dx = \lim_{t \to ∞} -e^{-x}\bigg|_{0}^{t} =$

$\lim_{t \to ∞} (-\frac{1}{e^t}+1) = 1 = 0!.$

Induction hypothesis, for k ≥ 0, the equality holds, $\int_{0}^{∞} x^ke^{-x}dx = k!$

Let’s consider $\int_{0}^{∞} x^{k+1}e^{-x}dx$ =[u = xk+1, du = (k+1)xkdx, dv = e-xdx, v = -e-x] = [1] + [2]

[1] $-x^{k+1}e^{-x}\bigg|_{0}^{∞} = 0 -0 = 0$. And the key is to understand that the exponent term e-x goes to zero much faster than any power of x grows to infinity.

Another way of seeing it, $\lim_{x \to ∞} \frac{x^k}{e^{x}} =$[L’Hôpital’s rule] $\lim_{x \to ∞} \frac{kx^{k-1}}{e^{x}} = \lim_{x \to ∞} \frac{k(k-1)x^{k-2}}{e^{x}} = ··· = \lim_{x \to ∞} \frac{some~constant}{e^{x}} = 0$

[2] $+ (k+1)\int_{0}^{∞} x^ke^{-x}dx =$[Induction hypothesis, for k ≥ 0, the equality holds, $\int_{0}^{∞} x^ke^{-x}dx = k!$] (k+1)k! = (k+1)!, hence $\int_{0}^{∞} x^ne^{-x}dx = n!$

# Arc lengths

In calculus, the concept of arc length refers to the length of a curve in a two-dimensional plane. The integral represents the sum of all the infinitesimally small line segments along the curve, and the square root term ensures that the Pythagorean theorem is applied to these segments, that is, (Δs)2 ≈ (Δx)2 + (Δy)2, and then, in the infinitesimal scale, (ds)2 = (dx)2 + (dy)2 ⇒ ds = $\sqrt{dx^2+dy^2} = dx\sqrt{1+(\frac{dy}{dx})^2}$ (Figure 2.a.).

Hence, our length is $\int_{s_0}^{s_n} ds = \int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^2}dx = \int_{a}^{b} \sqrt{1+f’(x)^2}dx$ where y = f(x).

• Let’s suppose y = mx, y’ = dy/dx = m, ds = $\sqrt{1+(y’)^2}dx = \sqrt{1+m^2}dx$, the length between 0 and 10 is $\int_{0}^{10} \sqrt{1+m^2}dx = 10\sqrt{1+m^2}dx$ (Figure 2.b.)

• Let’s assume y = $\sqrt{1-x^2}$ and calculate the arc length along the unit circle (Figure 2.c.), y’ = $\frac{-x}{\sqrt{1-x^2}}$, and the more complicated calculations implies ds = $\sqrt{1+(y’)^2}dx =$ [1+(y’)2 = $1+(\frac{-x}{\sqrt{1-x^2}})^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$] $\frac{dx}{\sqrt{1-x^2}}$.

The arc length is α = $\int_{0}^{a} \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x\bigg|_{a}^{0} = sin^{-1}a$ ↭ a = sin(α).

• Length of a parabola. y = x2, y’ = 2x, y’ = $\sqrt{1+(2x)^2}dx$ (Figure 2.d.).

The arc length (0 ≤ x ≤ a) is $\int_{0}^{a} \sqrt{1+4x^2}dx$.

Let’s calculate its surface area of the parabola y = x2 rotated around the x-axis, dA = (2πy)ds, hence its surface area = $\int_{0}^{a} (2πy)ds = \int_{0}^{a} (2πx^2)\sqrt{1+4x^2}dx$

• Surface area of a sphere with radius a. y = $\sqrt{a^2-x^2}, y’=\frac{-x}{\sqrt{a^2-x^2}}, 1 + (y’)^2 = 1 + \frac{x^2}{a^2-x^2} = \frac{a^2-x^2+x^2 }{a^2-x^2} = \frac{a^2}{a^2-x^2}$ (Figure 2.e.)

Area = $\int_{x_1}^{x_2} 2πyds = \int_{x_1}^{x_2} 2π\sqrt{a^2-x^2}\sqrt{\frac{a^2}{a^2-x^2}}dx = \int_{x_1}^{x_2} 2πadx = 2πa(x_2-x_1)$, e.g., the hemisphere x1 = 0, x2 = a, area = 2πa(a) = 2πa2; and the whole sphere, x1 = -a, x2 = a, area = 2πa(a -(-a)) = 4πa.

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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