Question

Find the area of the shaped region.

Short answer

The area of the shaded region is\(\frac{{16}}{3} - \frac{{4\sqrt 2 }}{3} - ln\left( 3 \right)\)

Step-by-step solution

Given information

The given equation is:

\(y = \sqrt {x + 2} \)

Finding the area of shaped region

Between \(x = 0\) and \(x = 2\)is the shaded zone.

\(f(x) = \sqrt {x + 2} \)

\(g(x) = \frac{1}{{x + 1}}\)

On the interval,\(f\)and\(g\)are continuous\(\left( {0,2} \right)\)

For all x in the range\(\left( {0,2} \right)\)is\(f(x) \ge g(x)\). The shaded area A of the region bordered by the lines x=0 and x=s, as well as the curves\(y = f(x)\)and\(y = g(x)\)

\(A = \int\limits_0^2 {\left( {\sqrt {x + 2} - \frac{1}{{x + 1}}} \right)} dx = \int\limits_0^2 {\sqrt {x + 2} } dx - \int\limits_0^2 {\frac{1}{{x + 1}}} dx\)

Using Substitution

use the substitution \(z = x + 1:dz = dx\)

when\(x = 0\)we have\(z = 0 + 1 = 1\)

when\(x = 2\)we have\(z = 2 + 1 = 3\)then use\(\int {\frac{1}{z}} dz = \ln (z)\)

\(\begin{aligned}{l}\int_2^4 {\sqrt u } du = \int_2^4 {{u^{\frac{1}{2}}}} du\\ = \left( {\frac{{{u^{\left( {\frac{1}{2} + 1} \right)}}}}{{\frac{1}{2} + 1}}} \right)\end{aligned}\)

\(\begin{aligned}{l} = \left( {\frac{2}{3}u\frac{3}{2}} \right)\\ = \frac{2}{3}{4^{\frac{3}{2}}} - \frac{2}{3}{2^{\frac{3}{2}}}\end{aligned}\)

\( = \frac{{16}}{3} - \frac{{4\sqrt 2 }}{3}\)

Use the substitution\(z = x + 7:dz = dx\)

when\(x = 0\)we have\(z = 0 + 1 = 1\)

when\(x = 2\)we have\(z = 2 + 1 = 3\)then use\(\int {\frac{1}{z}} dz = \ln (z)\)

\(\int_0^2 {\frac{1}{{x + 1}}} dx = \int_1^3 {\frac{1}{z}} dz\)

\(\begin{aligned}{l} = (\ln (z))\\ = \ln (3) - \ln (1)\\ = \ln (3)\end{aligned}\)

Finally, we calculate the area

\(\begin{aligned}{l}A = \int_0^2 {\sqrt {x + 2} } dx - \int_0^2 {\frac{1}{{x + 1}}} dx\\ = \frac{{16}}{3} - \frac{{4\sqrt 2 }}{3} - \ln (3)\end{aligned}\)