Question

find the area of the region in the first quadrant by bounded by the curve y=sqrt x^2+5, the x-axis and the line x=2

Answer 1

ANSWER: Area of region $\cong 5.0118$

EXPLANATION:

Let $D=\left \{ (x,y):0\leq x\leq2,\, 0\leq y\leq \sqrt{x^{2}+5} \right \} , A=Area$ of $D$ , then

$A=\int_{0}^{2}\sqrt{x^{2}+5}dx$

We calculate the integral by parts : denote $u= \sqrt{x^{2}+5} ,\, dv=dx\,$ . Therefore , $v=x, \, du=\frac{2x}{2\sqrt{x^{2}+5}}=\frac{ x}{ \sqrt{x^{2}+5}}$ and

$A=\left [ x\cdot \sqrt{x^{2}+5} \right ]_{0}^{2}-\int_{0}^{2}\frac{x^{2}}{\sqrt{x^{2}+5}}dx=\left [ 2\cdot \sqrt{2^{2}+5}-0 \right ] -\int_{0}^{2}\frac{x^{2}+5-5}{\sqrt{x^{2}+5}}dx=6-A+5\cdot \int_{0}^{2}\frac{1}{\sqrt{x^{2}+5}}dx=6-A+5\cdot \left [ \ln \left | x+\sqrt{x^{2}+5} \right | \right ]_{0}^{2}=6-A+5\cdot \left ( \ln (2+3) -\ln \sqrt{5}\right )=6-A+5\ln \frac{5}{\sqrt{5}}=6-A+5\cdot\frac{1}{2}\ln 5 .$

From the equality $A=6-A+5\cdot\frac{1}{2}\ln 5$ it follows $A=3+\frac{5}{4}\ln 5 \cong 5.0118.$