Calculation rules for integrals

3.3 Calculation rules for integrals

Nearly all integrals are solved by applying the fundamental theorem of calculus: \begin{eqnarray*}\int\limits_a^b f(x) dx&=& F(b)-F(a)\quad \mbox{where $f(x)=\frac{dF}{dx}$}\\ &=&\left[F(x)\right]_a^b\end{eqnarray*}

$\Rightarrow\;$ Calculation of integrals$\;\Leftrightarrow\;$Finding function $F(x)$
A function $F$ with the property $f(x)=\frac{dF}{dx}$ is called primitive with respect to $f(x)$.
Plus two rules:

integration by parts

\[\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x) dx\]

which directly follows from the product rule for calculating derivatives.
substitution:

\[\int f(g(x))g'(x) dx=F(g(x))+C\]

which directly follows from the chain rule for calculation derivatives.

Integrals: basic rules

\[ \begin{array}{ccl} \int\limits_a^b f(x)dx&=&-\int\limits_b^a f(x)dx\;\rightarrow\; \mbox{area with sign!}\\ \int\limits_a^af(x) dx&=&0\\ \int\limits_a^b\left[f(x)\pm g(x)\right] dx&=&\int\limits_a^b f(x) dx\pm\int\limits_a^b g(x) dx\\ \int\limits_a^b k f(x) dx&=&k\int\limits_a^b f(x) dx\quad\Rightarrow\;\mbox{''vector space''}\\ \int\limits_a^bf(x) dx&=&0\,\not\Rightarrow\,f(x)\equiv 0 \qquad \qquad \qquad \qquad \quad \mbox{because} \end{array} \]

Indefinite integral: \begin{eqnarray*}&&\int\limits_{x_0}^xf(y) dy=F(x)-F(x_0)\\ \rightarrow\;F(x)&=&\int\limits_{x_0}^x f(y) dy+C\\ \mbox{ or shorter}\;\;F(x)&=&\int f(x) dx+C\;\leftarrow\;\mbox{Indefinite integral, means } \int\limits_{x_0}^x f(y) dy!! \end{eqnarray*}

$\Rightarrow\;$ basic integrals: \begin{eqnarray*} \int x^n dx&=& \frac{1}{n+1} x^{n+1}+C, n\neq-1\, n\in\mathbb{Z}\\ \int x^r dx&=&\frac{1}{r+1} x^{r+1}+C,\,r\in\mathbb{R}\,\mbox{but } r\neq-1\\ \int \frac{1}{x} dx&=&\ln|x|+C\\ \int e^x dx &=& e^x +C\\ \int\sin x dx&=&-\cos x+C\\ \int\cos x dx&=&\sin x+C\end{eqnarray*}

Example for substitution rule:

\[\int\limits_0^1 x e^{x^2} dx=?\]

\[\begin{array}{lclcccl} g(x)&=&x^2&\rightarrow&g'(x)&=&2x\\ f(g)&=&e^g&\rightarrow&F(g)&=&e^g\\\\ &&&\rightarrow&\int xe^{x^2}dx&=&\int\frac{1}{2}g'(x)e^{g(x)} dx\\ &&&&&=&\frac{1}{2}e^{x^2}\\ &&&\rightarrow&\int\limits_0^1 xe^{x^2}dx&=&\left[\frac{1}{2}e^{x^2}\right]_0^1=\frac{1}{2}(e-1) \end{array} \]