Primitive function/Substitution/Section

Theorem

Suppose that ${}I$ denotes a real interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a continuous function. Let

$g\colon [a,b]\longrightarrow I$

be a continuously differentiable function. Then

${}\int _{a}^{b}f(g(t))g'(t)\,dt=\int _{g(a)}^{g(b)}f(s)\,ds\,$

holds.

Proof

Since ${}f$ is continuous and ${}g$ is continuously differentiable, both integrals exist. Let ${}F$ denote a primitive function for ${}f$, which exists, due to fact. Because of the chain rule, the composite function

${}t\mapsto F(g(t))=(F\circ g)(t)\,$

has the derivative ${}F'(g(t))g'(t)=f(g(t))g'(t)$. Therefore,

${}\int _{a}^{b}f(g(t))g'(t)\,dt=(F\circ g)|_{a}^{b}=F(g(b))-F(g(a))=F|_{g(a)}^{g(b)}=\int _{g(a)}^{g(b)}f(s)\,ds\,.$

$\Box$

Example

Typical examples, where one sees immediately that one can apply substitution, are

$\int g^{n}g',$

with the primitive function

${\frac {1}{n+1}}g^{n+1}$

or

$\int {\frac {g'}{g}},$

with the primitive function

$\ln g.$

Often, the indefinite integral is not in a form where one can apply the preceding rule directly. Then the following variant is more appropriate.

Corollary

Suppose that

$f\colon [a,b]\longrightarrow \mathbb {R}$

is a continuous function, and let

$\varphi \colon [c,d]\longrightarrow [a,b],s\longmapsto \varphi (s),$

denote a bijective continuously differentiable function. Then

${}\int _{a}^{b}f(t)\,dt=\int _{\varphi ^{-1}(a)}^{\varphi ^{-1}(b)}f(\varphi (s))\cdot \varphi '(s)\,ds\,$

holds.

Proof

Because of fact, we have

${}\int _{\varphi ^{-1}(a)}^{\varphi ^{-1}(b)}f(\varphi (s))\varphi '(s)\,ds=\int _{\varphi {\left(\varphi ^{-1}(a)\right)}}^{\varphi {\left(\varphi ^{-1}(b)\right)}}f(t)\,dt=\int _{a}^{b}f(t)\,dt\,.$

$\Box$

Remark

The substitution is applied in the following way: suppose that the integral

$\int _{a}^{b}f(t)\,dt$

has to be computed. Then one needs an idea that the integral gets simpler by the substitution

${}t=\varphi (s)\,$

(taking into account the derivative ${}\varphi '(s)$ and that the inverse function ${}\varphi ^{-1}$ has to be determined). Setting ${}c=\varphi ^{-1}(a)$ and ${}d=\varphi ^{-1}(b)$, we have the situation

$[c,d]{\stackrel {\varphi }{\longrightarrow }}[a,b]{\stackrel {f}{\longrightarrow }}\mathbb {R} .$

In certain cases, some standard substitutions help.

In order to make a substitution, three operations have to be done.

1. Replace ${}f(t)$ by ${}f(\varphi (s))$.
2. Replace ${}dt$ by ${}\varphi '(s)ds$.
3. Replace the integration bounds ${}a$ and ${}b$ by ${}\varphi ^{-1}(a)$ and ${}\varphi ^{-1}(b)$.

To remember the second step, think of

${}dt=d\varphi (s)=\varphi '(s)ds\,,$

which in the framework "differential forms“, has a meaning.

Example

The upper curve of the unit circle is the set

${\left\{(x,y)\mid x^{2}+y^{2}=1,\,-1\leq x\leq 1,\,y\geq 0\right\}}.$

For a given ${}x$, ${}-1\leq x\leq 1$, there exists exactly one ${}y$ fulfilling this condition, namely ${}y={\sqrt {1-x^{2}}}$. Hence, the area of the upper half of the unit circle is the area beneath the graph of the function ${}x\mapsto {\sqrt {1-x^{2}}}$, above the interval ${}[-1,1]$, that is

$\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx.$

Applying substitution with

$x=\cos t{\text{ and }}t=\arccos x$

(where ${}\cos :[0,\pi ]\rightarrow [-1,1]$ is bijective, due to fact), we obtain, using example, the identities

${}{\begin{aligned}\int _{a}^{b}{\sqrt {1-x^{2}}}\,dx&=\int _{\arccos a}^{\arccos b}{\sqrt {1-\cos ^{2}t}}(-\sin t)\,dt\\&=-\int _{\arccos a}^{\arccos b}\sin ^{2}t\,dt\\&={\frac {1}{2}}(\sin t\cos t-t)|_{\arccos a}^{\arccos b}.\end{aligned}}$

In particular, we get that

${}{\frac {1}{2}}{\left(x\cdot \sin {\left(\arccos x\right)}-\arccos x\right)}={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}\,$

is a primitive function for ${}{\sqrt {1-x^{2}}}$. Therefore,

${}{\begin{aligned}\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx&={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}|_{-1}^{1}\\&={\frac {1}{2}}(-\arccos 1+\arccos(-1))\\&=\pi /2.\end{aligned}}$