KindiakMath

Baby Integration

Let’s introduce baby integration as reverse-differentiation, which, comically, I have criticised elsewhere on my blog.

Given functions f(x) and F(x), write

\displaystyle \int f(x)\, \mathrm dx = F(x) \quad \iff \quad \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x).

We call the left-hand equation the indefinite integral of f(x) with respect to x.

(Wrong) Example 1. Recall that \displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = n x^{n-1}. Therefore,

\displaystyle \int n x^{n-1}\, \mathrm dx = x^n.

Now strictly speaking, our integrals are wrong.

Lemma 1. Suppose f(x) and F(x) are functions such that

\displaystyle \frac{\mathrm d}{ \mathrm dx}(F(x)) = f(x).

Then, for any real number C,

\displaystyle \frac{\mathrm d}{ \mathrm dx}(F(x) + C) = f(x).

Proof. Using the linearity of differentiation,

\begin{aligned} \frac{\mathrm d}{ \mathrm dx}(F(x) + C) &= \frac{\mathrm d}{ \mathrm dx}(F(x)) + \frac{\mathrm d}{ \mathrm dx}(C) \\ &= f(x) + 0 \\ &= f(x). \end{aligned}

Therefore, we use the following definition for indefinite integrals.

Definition 1. Given functions f(x) and F(x), write

\displaystyle \int f(x)\, \mathrm dx = F(x) +C \quad \iff \quad \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x),

where C is an arbitrary constant of integration.

Example 1. Recall that \displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = n x^{n-1}. Therefore, for n \neq 0,

\displaystyle \int n x^{n-1}\, \mathrm dx = x^n + C.

Now, Definition 1 is far from complete, but works well enough for most secondary-school assessments.

Theorem 1. Indefinite integration is linear: given functions f(x), g(x) and any real constant c,

\begin{aligned} \int(f(x) + g(x))\, \mathrm dx &= \int f(x)\, \mathrm dx + \int g(x)\, \mathrm dx, \\ \int cf(x)\, \mathrm dx &= c \int f(x)\, \mathrm dx. \end{aligned}

Proof. Write

\displaystyle \int f(x)\, \mathrm dx = F(x) + C_1,\quad \int g(x)\, \mathrm dx = G(x) + C_2

so that

\displaystyle \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) = f(x),\quad \frac{\mathrm d}{\mathrm dx}(G(x) + C_1) = g(x).

By the linearity of differentiation,

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(F(x) + C_1 + G(x) + C_2) &= \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) + \frac{\mathrm d}{\mathrm dx}(G(x) + C_2) \\ &= f(x) + g(x). \end{aligned}

For the second result, the linearity of differentiation yields

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(c\cdot (F(x) + C_1)) &= c \cdot \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) = c f(x). \end{aligned}

By Definition 1,

\begin{aligned} \int( c f(x) )\, \mathrm dx &= c \cdot ( F(x) + C_1 ) = c \int f(x)\, \mathrm dx . \end{aligned}

Example 2. For any n \neq -1, evaluate \displaystyle \int x^n\, \mathrm dx. In particular, for any constant a, evaluate

\displaystyle \int a\, \mathrm dx,\quad \int \sqrt x\, \mathrm dx,\quad \int \frac 1{x^2}\, \mathrm dx.

Solution. From Example 1, if m \neq 0, then

\displaystyle \int mx^{m-1}\, \mathrm dx = x^m + C.

Hence, writing m:=n+1 \iff n := m-1, for n \neq -1,

\displaystyle \int (n+1)x^n\, \mathrm dx = x^{n+1} + C.

By the linearity of integration,

\displaystyle (n+1)\int x^n\, \mathrm dx = x^{n+1} + C.

Therefore,

\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + \frac{C}{n+1}.

Since C' := C/(n+1) is still an arbitrary real constant,

\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + C'.

The other results follow from linearity and the law of exponents:

\begin{aligned} \int a\, \mathrm dx &= \int a x^0\, \mathrm dx = a \int x^0 \, \mathrm dx = a \cdot \frac x1 + C = ax + C, \\ \int \sqrt x\, \mathrm dx &= \int x^{1/2}\, \mathrm dx = \frac{x^{3/2}}{3/2} + C = \frac 23 \cdot x\sqrt x + C, \\ \int \frac 1{x^2}\, \mathrm dx &= \int x^{-2}\, \mathrm dx = \frac{x^{-1}}{-1} + C = -\frac 1x + C. \end{aligned}

Example 3. Evaluate \displaystyle \int (x^2 + 1)^2\, \mathrm dx.

Solution. Expanding (x^2 + 1)^2 yields

(x^2 + 1)^2 = (x^2)^2 + 2x^2 + 1^2 = x^4 + 2x^2 + 1.

Since integration is linear,

\begin{aligned} \int (x^2 + 1)^2 \, \mathrm dx &= \int (x^4 + 2x^2 + 1)\, \mathrm dx \\ &= \int x^4\, \mathrm dx + 2 \int x^2\, \mathrm dx + \int 1\, \mathrm dx \\ &= \frac{ x^5 }{ 5 } + 2 \cdot \frac{ x^3 }{ 3 } + x + C \\ &= \textstyle \frac 15 x^5 + \frac 23 x^3 + x + C. \end{aligned}

Remark 1. Where possible, it is good practice to present your final answer as a sum of terms of the form Ax^nf(x), where A,n are constants with n being a positive integer and f(x) is not a polynomial.

Example 3 should suggest that computing integrals exactly would require a lot more effort than computing derivatives exactly. That is correct. For instance, we don’t actually yet know how to calculate

\displaystyle \int \frac 1x\, \mathrm dx = \int x^{-1}\, \mathrm dx,

since Example 2 only works for n \neq -1.

Eventually, we will see that

\displaystyle \int \frac 1x\, \mathrm dx = \ln|x| + C,

which should surprise us—how did the logarithm magically appear there? For another fun result,

\displaystyle \int \frac 1{ 1 + x^2 }\, \mathrm dx = \tan^{-1}(x) + C.

What else could we do with integration?

Example 4. Let a, b be real constants. Evaluate \displaystyle \frac{\mathrm d}{\mathrm dx}((ax+b)^n). Hence, evaluate

\displaystyle \int (ax+b)^n\, \mathrm dx.

Solution. Using the chain rule,

\begin{aligned} \frac{\mathrm d}{\mathrm dx}((ax+b)^n) &= n(ax + b)^{n-1} \cdot \frac{\mathrm d}{\mathrm dx}(ax+b) \\ &= n(ax+b)^{n-1} \cdot \left(a \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(b) \right) \\ &= n(ax+b)^{n-1} \cdot (a\cdot 1 + 0) \\ &= an(ax+b)^{n-1}.\end{aligned}

By Definition 1 and the linearity of integration,

\begin{aligned}\int an(ax+b)^{n-1}\, \mathrm dx &= (ax+b)^n + C \\ an\int (ax+b)^{n-1}\, \mathrm dx &= (ax+b)^n + C \\\int (ax+b)^{n-1}\, \mathrm dx &= \frac {(ax+b)^n}{an} + \frac{C}{an} \\ &= \frac {(ax+b)^n}{an} + C', \end{aligned}

where C' := C/(an) is an arbitrary real constant of integration.

Using the same idea in Example 4, we have the following special form of integration.

Theorem 2. If \displaystyle \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x), then for real constants a, b,

\displaystyle \int f(ax+b)\, \mathrm dx = \frac 1a \cdot F(ax+b) + C.

Proof. Using the chain rule,

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(F(ax+b)) &= f(ax + b) \cdot \frac{\mathrm d}{\mathrm dx}(ax+b) \\ &= a \cdot f(ax+b).\end{aligned}

By Definition 1 and the linearity of integration,

\begin{aligned}\int a f(ax+b)\, \mathrm dx &= F(ax+b) + C \\ a \int f(ax+b)\, \mathrm dx &= (ax+b)^n + C \\ \int f(ax+b)\, \mathrm dx &= \frac {1}{a} \cdot F(ax+b) + \frac{C}{a} \\ &= \frac { 1 }{ a } \cdot F(ax+b) + C', \end{aligned}

where C' := C/a is an arbitrary real constant of integration.

Remark 1. Theorem 2 is a very special case of integrating by substitution, which is, in spirit, simply the reverse of the usual chain rule in differentiation.

Example 5. Differentiate x(x^2+1)^3. Hence, using Example 3, evaluate

\displaystyle \int (x^2+1)^3\, \mathrm dx.

Solution. Using the product rule and the chain rule,

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x(x^2+1)^3) &= \frac{\mathrm d}{\mathrm dx}(x) \cdot (x^2 + 1)^3 + \frac{\mathrm d}{\mathrm dx}((x^2+1)^3) \cdot x \\ &= 1 \cdot (x^2 + 1)^3 + 3(x^2+1)^2 \cdot 2x \cdot x \\ &= (x^2 + 1)^3 + 6x^2(x^2+1)^2 \\ &= (x^2 + 1)^3 + 6((x^2 + 1) - 1)(x^2+1)^2 \\ &= (x^2 + 1)^3 + 6((x^2+1)^3-(x^2+1)^2) \\ &= 7(x^2 + 1)^3 - 6(x^2+1)^2. \end{aligned}

Integrating on both sides and using linearity,

\begin{aligned} x(x^2 + 1)^3 + C_1 &= 7 \int(x^2+1)^3\, \mathrm dx - 6 \int (x^2 + 1)^2\, \mathrm dx.\end{aligned}

Doing algebra and using Example 3,

\begin{aligned} \int(x^2+1)^3\, \mathrm dx &= \frac 17 \cdot (x(x^2 + 1)^3 + C_1) + \frac 67 \int (x^2 + 1)^2\, \mathrm dx \\ &= \frac 17 \cdot (x(x^2 + 1)^3 + C_1) + \frac 67 \left( \frac 15 x^5 + \frac 23 x^3 + x + C_2 \right) \\ &= \textstyle \frac 17 \cdot x(x^2 + 1)^3 + \frac 6{35} x^5 + \frac 47 x^3 + \frac 67 x + C', \end{aligned}

where C' := \frac 17 C_1 + \frac 67 C_2 is an arbitrary real constant of integration.

Remark 2. Example 5 is a guided example of integrating by parts, which is, in spirit, simply the reverse of the usual product rule in differentiation.

There is a lot more to say about integration, especially since we have not even discussed calculus involving the non-algebraic functions like the trigonometric functions \sin(x), \cos(x), \tan(x), as well as the exponential function e^x and its inverse the natural logarithm \ln(x).

Nevertheless, we will push these discussions to the final post on integration, and pursue the essentials of the topic first. These functions have massive uses in the STEM fields, but for now we want to cover the big-idea bases of calculus that do not necessarily require them.

As such, we will apply integration to computing areas, as well as discuss some basic physics.

—Joel Kindiak, 18 Jan 25, 1200H

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