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Integration by parts II

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Recall

Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.

Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.

The process of finding antiderivatives is called integration.

The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$

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 -$\int \frac{1}{x}xdx = xlnx -x + C.$ 🚀

 

Solved exercises

 

The obvious 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 a previous exercise = $π(y(lny)^2-2(ylny -y)) \bigg|_{1}^{e} = π(e·1^2)- 2π(e·1-e) -π(1·0^2-2(1·0-1))= πe -0 -2π= π(e-2)$.

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 $=[A previous exercise]$ 2πe\frac{x^2}{2} -2πxe^x +2πe^x\bigg|_{0}^{1} = 2πe\frac{1}{2}-2πe+2πe-2π = πe -2π = π(e-2).$

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 $\int_{0}^{∞} x^ne^{-x}dx = n!$

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  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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