Solution 2.2:2c
From Förberedande kurs i matematik 2
If we focus on the integrand, then the substitution \displaystyle u=3x+1 seems suitable, since we then get \displaystyle \sqrt{u} which we can integrate. There is also no risk involved in using a linear substitution such as \displaystyle u=3x+1, because the relation between \displaystyle dx and \displaystyle du will be a constant factor,
| \displaystyle du = (3x+1)'\,dx = 3\,dx\,, |
which does not cause any problems.
We obtain
| \displaystyle \begin{align}
\int\limits_0^5 \sqrt{3x+1}\,dx &= \left\{\begin{align} u &= 3x+1\\[5pt] du &= 3\,dx \end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt] &= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt] &= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt] &= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.} \end{align} |
