question let the region r be the area enclosed by the function f(x) = √x + 1 and g(x) = x + 1. if the region…

Explanation:

Step1: Find intersection points

Set $\sqrt{x}+1=x + 1$. Subtract 1 from both sides: $\sqrt{x}=x$. Square both sides: $x=x^{2}$, which gives $x^{2}-x=0$, factoring out $x$ we get $x(x - 1)=0$. So the intersection points are $x = 0$ and $x=1$.

Step2: Determine side - length of square cross - section

The side - length $s$ of each square cross - section perpendicular to the $x$ - axis is $s=(\sqrt{x}+1)-(x + 1)=\sqrt{x}-x$.

Step3: Set up volume integral

The volume $V$ of the solid with square cross - sections is given by the integral $V=\int_{a}^{b}s^{2}dx$, where $a = 0$, $b = 1$ and $s=\sqrt{x}-x$. So $V=\int_{0}^{1}(\sqrt{x}-x)^{2}dx=\int_{0}^{1}(x - 2x^{\frac{3}{2}}+x^{2})dx$.

Step4: Integrate term by term

$\int_{0}^{1}(x - 2x^{\frac{3}{2}}+x^{2})dx=\left[\frac{1}{2}x^{2}-\frac{4}{5}x^{\frac{5}{2}}+\frac{1}{3}x^{3}\right]_{0}^{1}$.

Step5: Evaluate the definite integral

$\left(\frac{1}{2}(1)^{2}-\frac{4}{5}(1)^{\frac{5}{2}}+\frac{1}{3}(1)^{3}\right)-\left(\frac{1}{2}(0)^{2}-\frac{4}{5}(0)^{\frac{5}{2}}+\frac{1}{3}(0)^{3}\right)=\frac{1}{2}-\frac{4}{5}+\frac{1}{3}=\frac{15 - 24+10}{30}=\frac{1}{30}\approx0.033$.

Answer:

$0.033$