Let the slope of the line 45x+5y+3=0 be 27r_1+9r_22 for some r_1,r_2 R.

Question

Let the slope of the line $45x+5y+3=0$ be $27{r}_{1}+\frac{9{r}_{2}}{2}$ for some ${r}_{1},{r}_{2}\in R$. Then $\underset{x	o 3}{lim}({\int }_{x}^{3}\frac{8{t}^{2}}{\frac{3{r}_{2}x}{2}-{r}_{2}{x}^{2}-{r}_{1}{x}^{3}-3x}dt)$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

According to the question,

$\begin{gathered} 27{r}_{1}+\frac{9{r}_{2}}{2}=-9 \ \underset{x	o 3}{lim}\frac{\int _{3}^{x}8{t}^{2}dt}{\frac{3{r}_{2}x}{2}-{r}_{2}{x}^{2}-{r}_{1}{x}^{3}-3x} \ =\underset{x	o 3}{lim}\frac{8{x}^{2}}{\frac{3{r}_{2}^{2}}{2}-2{r}_{2}x-3{r}_{1}{x}^{2}-3} \ =\frac{72}{\frac{3{r}_{2}}{2}-6{r}_{2}-27{r}_{1}-3} \ =\frac{72}{-\frac{9{r}_{2}}{2}-27{r}_{1}-3} \ =\frac{72}{9-3}=12 \end{gathered}$

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