Question 512 Marks
Evaluate the following definite integrals:
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
Answer
View full question & answer→We know that $\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{C}$
Now,
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
$=\big[\log(\text{x}+7)\big]^3_{-2}$
$=\big[\log10-\log5]^3_{-2}$
$=\log\frac{10}{5}$ $\Big[\because\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big]$
$=\log2$
$\therefore\ \int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}=\log2$
Now,
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
$=\big[\log(\text{x}+7)\big]^3_{-2}$
$=\big[\log10-\log5]^3_{-2}$
$=\log\frac{10}{5}$ $\Big[\because\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big]$
$=\log2$
$\therefore\ \int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}=\log2$