Success is not final, failure is not fatal: It is the courage to continue that counts, Winston Churchill.
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 F_{n}(x) = $\int (lnx)^n dx$, F_{n}(x) = x(lnx)^{n} -nF_{n-1}(x). Hence, F_{0}(x) = $\int (lnx)^0 dx = x$, F_{1}(x) = x(lnx) -F_{0}(x) = xlnx -x + C🚀, F_{2}(x) = x(lnx)^{2} -2F_{1}(x) = x(lnx)^{2} -2(xlnx -x) + C 🚂
$\int x^ne^xdx$ =[u = x^{n}, u’ = nx^{n-1}, v’ = e^{x} ⇒ v = e^{x}] x^{n}e^{x} -$\int nx^{n-1}e^xdx$, so F_{n}(x) = $\int x^ne^xdx = x^ne^x -nF_{n-1}(x)$.
F_{0}(x) = $\int e^xdx = e^x + C.$ F_{1}(x) = $\int xe^xdx$ = xe^{x} -e^{x} + C, and so on…
The answer is -horizontal slicing- $\int_{1}^{e} πx^2dy$ =[y = e^{x}, 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^{-1}x, u’ = c, v = x²/2] = $\frac{x^2}{2}tan^{-1}(x) - \int \frac{x^2}{2}\frac{1}{1+x^2}dx = \frac{x^2}{2}tan^{-1}(x) - \frac{1}{2}\int \frac{x^2}{1+x^2}dx$ = [^{x²}⁄_{1+x²} = ^{x²+1-1}⁄_{1+x²} = 1-^{1}⁄_{1+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 = x^{k+1}, du = (k+1)x^{k}dx, dv = e^{-x}dx, 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 $\int_{0}^{∞} x^ne^{-x}dx = n!$
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).
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 = x^{2} rotated around the x-axis, dA = (2πy)ds, hence its surface area = $\int_{0}^{a} (2πy)ds = \int_{0}^{a} (2πx^2)\sqrt{1+4x^2}dx$
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 x_{1} = 0, x_{2} = a, area = 2πa(a) = 2πa^{2}; and the whole sphere, x_{1} = -a, x_{2} = a, area = 2πa(a -(-a)) = 4πa.