Solution 2.1:2b
From Förberedande kurs i matematik 2
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| - | {{ | + | There is no ready made standard formula for a primitive function to our integrand, but if we expand |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \int\limits_{-1}^{2} (x-2)(x+1)\,dx | ||
| + | &= \int\limits_{-1}^{2} (x^2+x-2x-2)\,dx\\[5pt] | ||
| + | &= \int\limits_{-1}^{2} (x^2-x-2)\,dx | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | and write the last integral as | ||
| + | |||
| + | {{Displayed math||<math>\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx</math>}} | ||
| + | |||
| + | we see that the integrand consists of three terms of the type <math>x^n</math> and we can directly write down a primitive function, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx | ||
| + | &= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt] | ||
| + | &= \frac{2^3}{3} - \frac{2^2}{2} - 2\cdot\frac{2}{1} - \Bigl(\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2\cdot\frac{(-1)}{1}\Bigr)\\[5pt] | ||
| + | &= \frac{8}{3} - \frac{4}{2} - 4 - \Bigl(-\frac{1}{3}-\frac{1}{2}+2\Bigr)\\[5pt] | ||
| + | &= \frac{16-12-24+2+3-12}{6}\\[5pt] | ||
| + | &= -\frac{27}{6}\\[5pt] | ||
| + | &= -\frac{9}{2}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
There is no ready made standard formula for a primitive function to our integrand, but if we expand
| \displaystyle \begin{align}
\int\limits_{-1}^{2} (x-2)(x+1)\,dx &= \int\limits_{-1}^{2} (x^2+x-2x-2)\,dx\\[5pt] &= \int\limits_{-1}^{2} (x^2-x-2)\,dx \end{align} |
and write the last integral as
| \displaystyle \int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx |
we see that the integrand consists of three terms of the type \displaystyle x^n and we can directly write down a primitive function,
| \displaystyle \begin{align}
\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx &= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt] &= \frac{2^3}{3} - \frac{2^2}{2} - 2\cdot\frac{2}{1} - \Bigl(\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2\cdot\frac{(-1)}{1}\Bigr)\\[5pt] &= \frac{8}{3} - \frac{4}{2} - 4 - \Bigl(-\frac{1}{3}-\frac{1}{2}+2\Bigr)\\[5pt] &= \frac{16-12-24+2+3-12}{6}\\[5pt] &= -\frac{27}{6}\\[5pt] &= -\frac{9}{2}\,\textrm{.} \end{align} |
