Solve the problem by integration.
-Under specified conditions, the time $t$ (in min) required to form $x$ grams of a substance during a chemical reaction is given by $t = \int \frac { d x } { ( 7 - x ) ( 2 - x ) }$. Find the equation relating $t$ and $x$ if $x = 0 g$ when $t = 0$ min.
A) $t = \frac { 1 } { 5 } \ln \left| \frac { 2 - x } { 7 - x } \right| - \frac { 1 } { 5 } \ln \frac { 2 } { 7 }$
B) $t = \frac { 1 } { 5 } \ln \left| \frac { 7 - x } { 2 - x } \right| + \frac { 1 } { 5 } \ln \frac { 7 } { 2 }$
C) $t = \frac { 1 } { 5 } \ln \left| \frac { 7 - x } { 2 - x } \right| - \frac { 1 } { 5 } \ln \frac { 7 } { 2 }$
D) $t = \frac { 1 } { 5 } \ln \left| \frac { 2 - x } { 7 - x } \right| + \frac { 1 } { 5 } \ln \frac { 2 } { 7 }$

Question

Quizplus · Accepted Answer

Using partial fractions, we have$$\frac{1}{(7-x)(2-x)}=\frac{1}{5}\left(\frac{1}{2-x}-\frac{1}{7-x}\right)$$So,$$t=\int\frac{1}{(7-x)(2-x)}dx=\frac{1}{5}\int\left(\frac{1}{2-x}-\frac{1}{7-x}\right)dx$$$$=\frac{1}{5}\left(\ln|2-x|-\ln|7-x|\right)+C$$$$=\frac{1}{5}\ln\left|\frac{2-x}{7-x}\right|+C$$Using the initial condition, we get$$0=\frac{1}{5}\ln\left|\frac{2-0}{7-0}\right|+C$$$$0=\frac{1}{5}\ln\frac{2}{7}+C$$$$C=-\frac{1}{5}\ln\frac{2}{7}$$Therefore,$$t=\frac{1}{5}\ln\left|\frac{2-x}{7-x}\right|-\frac{1}{5}\ln\frac{2}{7}$$

Solve the Problem by Integration ttt (In Min) Required to Form xxx

Solve the Problem by Integration $t$ (In Min) Required to Form $x$