evaluate $\\int_{4}^{5}\\int_{1}^{2}(3x + y)^{-2}dydx$.
evaluate $\\int_{4}^{5}\\int_{1}^{2}(3x + y)^{-2}dydx$.
Answer
Explanation:
Step1: Integrate with respect to ( y )
Use the power - rule for integration (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). Let (u = 3x + y), then (du=dy). [ \begin{align*} \int_{1}^{2}(3x + y)^{-2}dy&=\left[\frac{(3x + y)^{-2 + 1}}{-2+1}\right]{y = 1}^{y = 2}\ &=\left[-\frac{1}{3x + y}\right]{y = 1}^{y = 2}\ &=-\frac{1}{3x+2}+\frac{1}{3x + 1} \end{align*} ]
Step2: Integrate the result with respect to ( x )
[ \begin{align*} \int_{4}^{5}\left(-\frac{1}{3x+2}+\frac{1}{3x + 1}\right)dx&=-\frac{1}{3}\int_{4}^{5}\frac{3}{3x+2}dx+\frac{1}{3}\int_{4}^{5}\frac{3}{3x + 1}dx\ \end{align*} ] Let (u_1=3x + 2), (du_1 = 3dx); (u_2=3x + 1), (du_2=3dx). When (x = 4), (u_1=14), (u_2 = 13); when (x = 5), (u_1=17), (u_2=16). [ \begin{align*} &=-\frac{1}{3}\left[\ln|3x+2|\right]{4}^{5}+\frac{1}{3}\left[\ln|3x + 1|\right]{4}^{5}\ &=-\frac{1}{3}(\ln17-\ln14)+\frac{1}{3}(\ln16-\ln13)\ &=\frac{1}{3}\left(\ln16-\ln13-\ln17+\ln14\right)\ &=\frac{1}{3}\ln\frac{16\times14}{13\times17}\ &=\frac{1}{3}\ln\frac{224}{221} \end{align*} ]
Answer:
(\frac{1}{3}\ln\frac{224}{221})