**3.7 Pengamiran, SPM Praktis (Kertas 1) **

**Soalan 2:**

Diberi bahawa $\int \frac{4}{{\left(1+x\right)}^{4}}dx=m{\left(1+x\right)}^{n}}+c,$

Cari nilai-nilai *m*dan
*n*.

*Penyelesaian:*

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

Diberi

${\int}_{-1}^{2}2g(x)dx=4}\text{,dan}{\displaystyle {\int}_{-1}^{2}\left[mx+3g\left(x\right)\right]dx}=15.$

Cari nilai pemalar *m*.

*Penyelesaian:*

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

Diberi

$\frac{d}{dx}\left(\frac{2x}{3-x}\right)=g\left(x\right)\text{,cari}{\displaystyle {\int}_{1}^{2}g(x)dx.}$

*Penyelesaian:*

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