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4.7 Integration techniques

Due to the Fundamental Theorem of Calculus (FTC), one may consider the process of integration for continuous functions as the inverse process of derivation

image​

Using this relations, we obtain integration techniques from the basic properties of differentiation.

Let us start with an example: let \(F, G\) be two differentiable functions on an interval \([a, b]\).

Recall that a rational function is a function of the form \(f(x) = \frac{p(x)}{q(x)}\)

where \(p(x)\) and \(q(x)\) are polynomials. The domain of \(f\) is \(Dom (f) = \R -\{\text{zeros of } q(x)\}\).

In its domain, \(f\) is a smooth function. In particular, it is integrable in its domain.

We know that \(f(x)=\frac{1}{1+x}\) and \(\displaystyle \int^x_0\frac{1}{1+t}dt=\ln(1+x)\).

Also \(\displaystyle \int^{1+x}_1\frac{1}{h}=\ln(x)\) where \(h=1+t\)

Example

\(f(x)=\frac{1}{x^{2}-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}\), then NT find \(A,B\in \R\) and \(x\neq 2,3\)​

Since \(\frac{A}{x-2}+\frac{B}{x-3}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}=\frac{1}{(x-2)(x-3)}\), then \(A(x-3)+B(x-2)=1\), then \(A=-1,B=1\)

Thus \(f(x)=-\frac{1}{x-2}+\frac{1}{x+3}\), then \(\displaystyle\int^{5}_{4}\frac{1}{x^{2}-5x+6}=\int^{5}_{4}-\frac{1}{x-2}+\frac{1}{x-3} dx=-\ln(x-2)|^5_4+\ln(x-3)|^5_4\)​

Theorem (Decomposition of real rational functions)

Let \(f (x) = \frac{p(x)}{q(x)}\) be a rational function with \(deg(p(x)) < deg(q(x))\) and assume that

\(q(x)=(x-a_{1})^{m_1}(x-a_{2})^{m_2}...(x-a_{\ell})^{m_{\ell}}q_{1}(x)^{r_1}q_{2} (x)^{r_2}...q_{s}(x)^{r_{s}}\)

Then there exists real numbers \(A_{11},A_{12},...,A_{1m_1},A_{21},A_{22},...,A_{2m_2},\ldots,A_{\ell1},A_{\ell2} ,...,A_{\ell m_{\ell}},B_{11},C_{11},B_{12},C_{12},\ldots\ldots,B_{1r_1},C_{1r_1} ,\ldots\ldots,B_{s1},C_{s1},\ldots,B_{sr_s},C_{sr_s}\) such that \(\frac{p(x)}{q(x)}=\frac{A_{11}}{(x-a_{1})}+\frac{A_{12}}{(x-a_{1})^{2}}+...+\frac{A_{1m_1}}{(x-a_{1})^{m_1}} +...+\frac{A_{\ell1}}{(x-a_{\ell})}+\frac{A_{\ell2}}{(x-a_{\ell})^{2}}+...+\frac{A_{\ell m_{\ell}}}{(x-a_{\ell})^{m_{\ell}}}+...+\frac{B_{11}x+C_{11}}{q_{1}(x)}+\frac{B_{12}x+C_{12}}{q_{1}(x)^{2}} +...+\frac{B_{1r_1}x+C_{1r_1}}{q_{1}(x)^{r_1}}+\) \(... + \frac{B_{s1}x + C_{s1}}{q_{s}(x)}+ \frac{B_{s2}x + C_{s2}}{q_{s}(x)^{2}}+ ... + \frac{B_{sr_s}x + C_{sr_s}}{q_{s}(x)^{r_s}}\)

Proof. (Idea) One looks for the numbers that solve the equality by using the sum of fractions and the equality of polynomials. Then one gets a linear system as in the example that always has a solution.

Remark

Let \(f(x) = \frac{p(x)}{q(x)}\) be a rational function.

By using Euclides Algorithm for the division of polynomials we may write \(p(x) = t(x) q(x) + r(x)\) with \(r(x) = 0\) or \(deg(r(x)) < deg(q(x))\)

Hence, \(f(x) = t(x) + \frac{r(x)}{q(x)}\)

So, for \([a, b] \subseteq Dom(f)\) we have that \(\displaystyle\int_{a}^{b} f(x)dx = \int_{a}^{b} t(x)dx + \int_{a}^{b} \frac{r(x)}{q(x)}dx\)

To solve the second term in right hand side, we apply the method above and substitution (see below)

Let us now draw our attention to another property of differentiation.

Let \(F, G\) be two differentiable functions on an interval \([a, b]\) and consider the function given by the product \((FG)(x) = F(x)G(x) ∀ x ∈ [a, b]\).

Then \((FG)'(x) = F'(x)G(x) + F(x)G'(x) ∀x ∈ [a, b]\)

Then we integrating on both sides, we get \({{\displaystyle\int_{a}^{x}\left(FG)^{\prime}(t\right)dt=\int_{a}^{x}F^{\prime}(t)G(t)+F\left(t\right)G^{\prime}\left(t\right)dt}}\)

Then \((FG)(x)-(FG)(a)=\displaystyle\int_{a}^{x}F^{\prime}(t)G(t)+\int_{a}^{x}F\left(t\right)G^{\prime} \left(t\right )dt\)​

\({\displaystyle\int_{a}^{x}F(t)G^{\prime}(t)dt=(FG)(t)|_{a}^{x}-\int_{a}^{x}F^{\prime}(t)G(t)dt}\)

Taking \(b = x\) and \(F(x) = u(x), G(x) = v(x)\), we get the formula for integration by parts:

Theorem (Integration by parts)

Let \(u, v\) be two differentiable functions on an interval \([a, b]\) such that \(u'\) and \(v'\) are integrable on \([a, b]\).

Then \(\displaystyle\int_{a}^{b} u(x)v'(x) dx = u(x)v(x) |_{a}^{b} - \int_{a}^{b} u'(x)v(x) dx\)

Proof. It follows from the arguments above

Example

  1. \(\displaystyle\int^\pi_0x\sin(x)dx\)

    Let \(u(x)=x\) since \(u(x)=1\) is easy and \(v'(x)=\sin x\), then \(v(x)=-\cos x\)

    Then \(\displaystyle\int_0^{\pi}\underbrace{x}_{u}\underbrace{\sin(x)}_{v^{\prime}}dx= x(-\cos x)|^\pi_0-\int^\pi_01(-\cos x)dx=-x\cos x|^\pi_0+\int^\pi_0\cos xdx\)

    \(=-\pi\cos(\pi)+0\cdot \cos(0)+\sin(x)|^{\pi}_0=-\pi(-1)+\sin(\pi)-\sin(0)=\pi\)

  2. \(\displaystyle \int^e_1\ln(x)dx\)

    Here since we don't know the primitive of \(\ln(x)\), thus we have to set \(u(x)=\ln(x)\)

    Let \(u(x)=\ln(x)\rightarrow u'(x)=\frac{1}{x}\) and \(v'(x)=1\rightarrow v(x)=x\)​

    Then \(\displaystyle \int^{e}_{1}\ln(x)dx=x\ln x|^e_1-\int^e_1\frac{1}{x}xdx=e\ln e-1\ln 1-\int^e_11dx=1\)

Finally, we end the lecture with the relation with the chain rule in differentiation.

Let \(F, G\) be two differentiable functions and assume that \(Im G(x) ⊆ Dom F(x)\), in particular, we may compose them and obtain the new function \((F ∘ G)(x) = F(G(x))\)

Then, \((F ∘ G)'(x) = F'(G(x))G'(x)\)

If we integrate on both sides with respect to a variable \(t\) in a closed interval \([a, b] ⊆ Dom (F)\) we get \({\displaystyle\int_{a}^{x}(F\circ G)^{\prime}(t)dt=\int_{a}^{x}F^{\prime}(G(t))G^{\prime}(t)dt}\)

Then by FTC, we get \({\displaystyle\int_{a}^{x}(F\circ G)^{\prime}(t)dt=(F\circ G)(x)-(F\circ G)(a)=F(G(x))-F(G(a))=(F\circ G)(t)|_{a}^{x}}\)

What can be said about the right hand side?

If \(G\) is injective, for example if \(G'(x) > 0\) for all \(x ∈ [a, b]\), we may consider \(G(x) = u(x)\) as a change of variable. Write \(F'(t) = f(t)\)

image​

Example

\(\displaystyle \int^4_3\frac{1}{x-2}dx\), but we know \(\displaystyle \int\frac{1}{t}dt=\ln(t)\)

Then we use change of variable, let \(u(x)=x-2=t\), then image​

Theorem (Substitution)

Suppose \(u: [c, d] → [a, b]\) is an injective differentiable function such that \(u'\) is integrable on \([c, d]\) and \(u(c) = a, u(d) = b\)

If \(f\) is a continuous function on \([a, b]\), then \(\displaystyle\int_{a}^{b} f(t) dt = \int_{c}^{d} f(u(x))u'(x) dx\)

Proof

Since \(f\) is continuous, then it is integrable. Also since \(u'\) is integrable, the function \(f(u(x))u'(x)\)is integrable because \(f(u(x))\) is continuous (exercise: the product of two integrable function is integrable. Idea: bounded one function)

Now we have to prove that both intervals are the same

Set \(F(x)=\displaystyle\int^x_{a}f(t)dt\), then \(F\) is a primitive of \(f\), by FTC, \(F'(x)=f(x)\)

So \(\displaystyle \int^{b}_{a}f(t)dt=F(b)-F(a)=F(u(d))-F(u(c))\) on the lefthand side

Now \(G(x)=F(u(x)),\forall x\in[c,d]\)

Then \(G'(x)=F'(u(x))\cdot u'(x)=f(u(x))\cdot u'(x),\forall x\in [c,d]\)

Hence, \(\displaystyle\int_{c}^{d}f(u(x))u'(x)dx=\int_{c}^{d}G^{\prime}(x)dx=G(d)-G(c)=F( u(d)) -F(u(c))\)​

Thus \(\displaystyle\int_{a}^{b} f(t) dt = \int_{c}^{d} f(u(x))u'(x) dx\)

Note: In some book one may find the notation \(\begin{cases} u(x)=t \\ u^{\prime}(x)dx=dt \end{cases}\) where \(u'(x)\) measure how the intervals change

The main idea of this theorem is to apply it to compute the integral on the right using that we might know a primitive of the function on the left.

The key step is to identify the function \(u\) and its derivative.

Exercise

If \(u'(x)\) and \(f(u(x))\) is integrable, the function \(f(u(x))u'(x)\)is integrable

Proof

Since \(u'(x)\) and \(f(u(x))\) is integrable, then \(u'(x),f(u(x))\) bounded.

Then \(m_{u'}<u'(x)<M_{u'}\) and \(m_{f\circ u}<f(u(x))<M_{f\circ u}\)​

Then \(U(\left(f\circ u)\cdot u^{\prime},P\right)-L(\left(f\circ u)\cdot u^{\prime},P\right )=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)\left(f\left(u\left(M_{i}\right)\right) \cdot u^{\prime}\left(M_{i}\right)-f\left(u\left(m_{i}\right)\right)\cdot u^{\prime} \left(m_{i}\right)\right)\)

We need to bound \(f\left(u\left(M_{i}\right)\right)\cdot u^{\prime}\left(M_{i}\right)-f\left(u\left (m_{i}\right)\right)\cdot u^{\prime}\left(m_{i}\right)=f\left(u\left(M_{i}\right) \right)\cdot u^{\prime}\left(M_{i}\right)+f\left(u\left(M_{i}\right)\right)\cdot u^{\prime}\left(m_{i}\right)-f\left(u\left(M_{i}\right)\right)\cdot u^{\prime}\left (m_{i}\right)-f\left(u\left(m_{i}\right)\right)\cdot u^{\prime}\left(m_{i}\right)\)

\(=f\left(u\left(M_{i}\right)\right)\cdot\left\lbrack u^{\prime}\left(M_{i}\right) -u^{\prime}\left(m_{i}\right)\right\rbrack+u^{\prime}\left(m_{i}\right)\cdot\left \lbrack f\left(u\left(M_{i}\right)\right)-f\left(u\left(m_{i}\right)\right)\right \rbrack\)

\(\leq M_{f\circ u}\cdot\varepsilon_{u^{\prime}}+M_{u^{\prime}}\cdot\varepsilon_{f\circ u}\)

......

Example

Compute \(\displaystyle \int^2_1\frac{1}{\sqrt{3x-2}}dx\)

Let \(u(x)=3x-2=t\), then \(u(1)=1,u(2)=4\), \(u'(x)=3\) and \(u'(x)dx=dt\), then \(3dx=dt\)

Then \(\displaystyle\int_1^2\frac{1}{\sqrt{3x-2}}dx=\frac{1}{3}\int_1^4\frac{1}{\sqrt{t}} dt=\frac{1}{3}\frac{t^\frac{3}{2}}{\frac{3}{2}}|^4_1=\frac{16}{9}-2\)

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