example 5.2\nevaluate $int_{1}^{4}\frac{dx}{(3 - 5x)^{2}}$

Explanation:

Step1: Use substitution

Let $u = 3 - 5x$, then $du=-5dx$. When $x = 1$, $u=3 - 5\times1=-2$; when $x = 4$, $u=3-5\times4=-17$. And $dx=-\frac{1}{5}du$.

Step2: Rewrite the integral

The integral $\int_{1}^{4}\frac{dx}{(3 - 5x)^2}$ becomes $\int_{-2}^{-17}\frac{-\frac{1}{5}du}{u^2}=-\frac{1}{5}\int_{-2}^{-17}u^{-2}du$.

Step3: Integrate $u^{-2}$

Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-2$, we have $\int u^{-2}du=-\frac{1}{u}+C$.

Step4: Evaluate the definite integral

$-\frac{1}{5}\left[-\frac{1}{u}\right]_{-2}^{-17}=-\frac{1}{5}\left(\frac{1}{-17}-\frac{1}{-2}\right)=-\frac{1}{5}\left(-\frac{1}{17}+\frac{1}{2}\right)=-\frac{1}{5}\times\frac{-2 + 17}{34}=-\frac{1}{5}\times\frac{15}{34}=-\frac{3}{34}$.

Answer:

$-\frac{3}{34}$