Mathematics 9709/22 — May/June 2013

Walkthrough

We start with the derivative $\frac{dy}{dx} = \frac{4}{7-2x}$. To find $y$, we integrate both sides with respect to $x$. The integrand is a constant multiple of $\frac{1}{7-2x}$. The standard result $\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + c$ applies directly. Here $a = -2$, so dividing the constant $4$ by $-2$ gives $-2$. Thus the indefinite integral is $-2\ln|7-2x| + C$.

Next, the point $(3,2)$ lies on the curve, so we substitute $x=3$ and $y=2$ into $y = -2\ln|7-2x| + C$. This gives $2 = -2\ln(7-6) + C = -2\ln(1) + C$. Since $\ln(1)=0$, we find $C = 2$. Therefore the equation of the curve is $y = -2\ln(7-2x) + 2$.

Key Takeaways

• To reverse a derivative, we integrate. A rational integrand of the form $\frac{k}{ax+b}$ integrates to $\frac{k}{a}\ln|ax+b|$ plus a constant.
• The constant of integration is found when we are given a specific point on the curve.
• Knowing that $\ln(1) = 0$ often simplifies evaluation.

Common Mistakes

• Forgetting the factor $\frac{1}{a}$ when integrating $\frac{1}{ax+b}$. Many students incorrectly write $\ln(7-2x)$ without the $-2$ factor.
• Omitting the constant of integration when performing an indefinite integral.
• Substituting the point incorrectly, or mixing up $x$ and $y$ coordinates.

Things to Be Careful About

• The argument of the logarithm is $7-2x$. At the given point $x=3$, it is $1$, which is positive, so we may write $\ln(7-2x)$ without the absolute value. However, in general the absolute value is needed unless the sign is known.
• Check the derivative of the final answer: $\frac{d}{dx}(-2\ln(7-2x)) = -2 \cdot \frac{-2}{7-2x} = \frac{4}{7-2x}$, confirming the integration is correct.