Antidifferentiation (Part 2: Indefinite Integration)

So last time we began discussing antiderivatives a bit. Lets continue discussing it more in this post.

Example 1. Consider
(1)\displaystyle  f(x)=\sin(x).
What is its antiderivative?

We recall the derivative of trigonometric functions tells us
(2)\displaystyle  \frac{\mathrm{d}}{\mathrm{d} x}\cos(x)=-\sin(x).
This tells us that
(3)\displaystyle  F(x)=-\cos(x)+C
where {C} is a constant.

Example 2. Take
(4)\displaystyle  g(x)=\cos(7x)+\sin(2x).
What is its antiderivative?

We see that
(5)\displaystyle  G(x)=\frac{\sin(7x)}{7}-\frac{\cos(2x)}{2}+C
is its antiderivative, for some constant {C}. We see this by differentiating {G'(x)=g(x)}.
(End of Example)

We can deduce several formulas for antiderivatives:

  1. If {f(x)=x^{n}}, then its antiderivative {\displaystyle F(x)=\frac{x^{n+1}}{n+1}+C}
  2. For {f(x)=1/x}, then {F(x)=\ln|x|+C} is its antiderivative;
  3. When {f(x)=a^{kx}=\mathrm{e}^{kx\ln(a)}}, then {\displaystyle F(x)=\frac{\mathrm{e}^{kx\ln(a)}}{k\ln(a)}+C};
  4. For {f(x)=0}, then {F(x)=C}.

These formulas are purely “rules of thumb” we observe from our examples.

However, we can invent some formal rules due to the properties of the derivative.

  1. If {g(x)=k f(x)} for some constant {k}, then the antiderivatives are {G(x)=k F(x)+C}.
  2. Let {h(x)=af(x)+bg(x)}, then the antiderivatives are {H(x)=aF(x)+bG(x)+C}.

Can we determine the constant {C}? We need more information to fix it to some value. But when asking specifically for the antiderivative, the constant {C} remains arbitrary. What sort of extra information do we need? Well, we need an “initial value problem”.

Example 3 (Typical Initial Value Problem). Given the equation
(6)\displaystyle  \frac{\mathrm{d} y}{\mathrm{d} x}=f(x)
determine {y(x)} such that it solves this equation, with the additional “initial” condition {y(x_{0})=y_{0}} (where {y_{0}}, {x_{0}} are constants).

If {F(x)} is the antiderivative of {f(x)} in Eq (6), then
(7)\displaystyle  y(x)=F(x)+C
is the solution. Is this all? Well, we can do more: we can fix {C}. How? Plug in {x_{0}} for {x}, obtaining
(8)\displaystyle  y(x_{0})=F(x_{0})+C=y_{0}
implies
(9)\displaystyle  C=y_{0}-F(x_{0}).
This solves the initial value problem.

Remark. We have seen an initial value problem once before, but phrased differently. See how we constructed the exponential function for details.

We introduce some notation and terminology. We define the “Indefinite Integral” of {f(x)} to be the function whose derivative is {f(x)}, denoted
(10)\displaystyle  \int f(x) \;\mathrm{d}{x} =\int(\mathrm{integrand})\;\mathrm{d}\!\begin{pmatrix}\mathrm{variable\ of}\\ \mathrm{integration} \end{pmatrix}
Some terminology: we call the “{\int}” the integral sign, the function {f(x)} the integrand, the {x} in {\mathrm{d}{x}} the variable of integration. We have
(11)\displaystyle  F(x)=\int f(x)\;\mathrm{d}{x}
in general.

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
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2 Responses to Antidifferentiation (Part 2: Indefinite Integration)

  1. Pingback: [Index] Calculus of a Single Variable | My Math Blog

  2. Pingback: [Index] Calculus in a Single Variable | My Math Blog

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