Definite Integral

Tags

Calculus

Cegep/2

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588 words

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4 minutes

Area under a curve between two boundaries

The definite integral of $f$ on $[a,b]$ is

$${\int }_a^{b}f(x)dx=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(x_i)\mathrm{\Delta }x$$

Derived from Riemann sum

Properties

• ${\int }_a^{b}kdx=k(b-a)$
• ${\int }_a^{a}f(x)dx=0$
• ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$
• ${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$
• Integration maintains inequality:
  • $f(x)\ge 0⟹{\int }_a^{b}f(x)dx\ge 0$
  • $f(x)\ge g(x)⟹{\int }_a^{b}f(x)dx\ge {\int }_a^{b}g(x)dx$
  • $m\le f(x)\le M⟹m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$
• For an even function $f$, ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$
• For an odd function $f$, ${\int }_{-a}^{a}f(x)dx=0$

Examples

Suppose ${\int }_0^{1}f(x)dx=2$, ${\int }_1^{2}f(x)dx=3$, ${\int }_0^{1}g(x)dx=-1$, and ${\int }_0^{2}g(x)dx=4$. Compute the following.

${\int }_1^{2}g(x)dx$

$$\begin{array}{rl}{\int }_1^{2}g(x)dx& ={\int }_0^{2}g(x)dx-{\int }_0^{1}g(x)dx\\ & =4+1\\ & =5\end{array}$$

${\int }_0^2(2f(x)-3g(x))dx$

$$\begin{array}{rl}{\int }_0^2(2f(x)-3g(x))dx& =2{\int }_0^{2}f(x)dx-3{\int }_0^{2}g(x)dx\\ & =2({\int }_0^{1}f(x)dx+{\int }_1^{2}f(x)dx)-3\cdot 4\\ & =2(2+3)-12\\ & =-2\end{array}$$

${\int }_1^{1}g(x)dx$

$${\int }_1^{1}g(x)dx=0$$

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Simplify the following expressions.

${\int }_{-2}^{2}f(x)dx+{\int }_2^{5}f(x)dx-{\int }_{-2}^{-1}f(x)dx$

$${\int }_{-2}^{2}f(x)dx+{\int }_2^{5}f(x)dx-{\int }_{-2}^{-1}f(x)dx={\int }_{-1}^{5}f(x)dx$$

${\int }_0^{2}3f(x)dx+{\int }_1^{3}3f(x)dx-{\int }_0^{2}2f(x)dx-{\int }_1^{2}3f(x)dx-{\int }_2^{3}2f(x)dx$

$$\begin{array}{rl}& {\int }_0^{2}3f(x)dx+{\int }_1^{3}3f(x)dx-{\int }_0^{2}2f(x)dx-{\int }_1^{2}3f(x)dx-{\int }_2^{3}2f(x)dx\\ =& {\int }_0^{2}3f(x)dx+{\int }_2^{3}3f(x)dx-{\int }_0^{3}2f(x)dx\\ =& 3{\int }_0^{3}f(x)dx-2{\int }_0^{3}f(x)dx\\ =& {\int }_0^{3}f(x)dx\end{array}$$

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Let $f(x)$ be an even function such that ${\int }_0^{2}3f(x)dx=6$ and ${\int }_1^{2}f(x)dx=5$. Find ${\int }_{-2}^{1}f(x)dx$.

$$\begin{array}{rl}{\int }_0^{2}3f(x)dx& =6\\ 2{\int }_0^{2}f(x)dx& =4\\ {\int }_{-2}^{2}f(x)dx& =4\\ {\int }_{-2}^{1}f(x)dx& =-1\end{array}$$

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Show that $\frac{1}{17}\le {\int }_1^2\frac{1}{1+x^4}\mathrm{d}x\le \frac{1}{2}$.

$$\begin{array}{rlrl}\frac{1}{1+2^4}& \le \frac{1}{1+x^4}& \le \frac{1}{1+1^4}& \text{ on }[1,2]\\ ⟹\frac{1}{17}(2-1)& \le {\int }_1^2\frac{1}{1+x^4}\mathrm{d}x& \le \frac{1}{2}(2-1)\end{array}$$

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Find a definite integral that represents $\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n\frac{2}{n}{e}^{(\frac{4i^2}{n^2}+\frac{4i}{n}+1)}$.

$$\begin{array}{r}{\int }_0^2{e}^{(x+1{)}^2}\mathrm{d}x\end{array}$$

Contributors

Ajitani Hifumi

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