Bab 14 Pengamiran

$\begin{array}{l}{\displaystyle \int \frac{4}{{\left(1+x\right)}^4}dx=m{\left(1+x\right)}^n}+c\\ {\displaystyle \int 4}{\left(1+x\right)}^{-4}dx=m{\left(1+x\right)}^n+c\\ \frac{4{\left(1+x\right)}^{-3}}{-3\left(1\right)}+c=m{\left(1+x\right)}^n+c\\ -\frac{4}{3}{\left(1+x\right)}^{-3}+c=m{\left(1+x\right)}^n+c\\ m=-\frac{4}{3},n=-3\end{array}$

$\begin{array}{l}{\displaystyle {\int }_{-1}^2\left[mx+3g\left(x\right)\right]dx}=15\\ {\displaystyle {\int }_{-1}^{2}mxdx}+{\displaystyle {\int }_{-1}^{2}3g\left(x\right)dx}=15\\ {\left[\frac{m{x}^2}{2}\right]}_{-1}^2+3{\displaystyle {\int }_{-1}^{2}g\left(x\right)dx}=15\\ \left[\frac{m{\left(2\right)}^2}{2}-\frac{m{\left(-1\right)}^2}{2}\right]+\frac{3}{2}{\displaystyle {\int }_{-1}^{2}2g\left(x\right)dx}=15\\ 2m-\frac{1}{2}m+\frac{3}{2}\left(4\right)=15\leftarrow \boxed{\text{diberi }{\displaystyle {\int }_{-1}^{2}2g(x)dx=4}}\\ \frac{3}{2}m+6=15\\ \frac{3}{2}m=9\\ m=9\times \frac{2}{3}\\ m=6\end{array}$

$\begin{array}{l}\text{Diberi }\frac{d}{dx}\left(\frac{2x}{3-x}\right)=g\left(x\right)\\ {\displaystyle \int g\left(x\right)dx=}\frac{2x}{3-x}\\ \text{dengan itu,}\\ {\displaystyle {\int }_1^{2}g(x)dx}={\left[\frac{2x}{3-x}\right]}_1^2\\ =\frac{2\left(2\right)}{3-2}-\frac{2\left(1\right)}{3-1}\\ =4-1\\ =3\end{array}$