## Task Condition

Calculate the area of the figure bounded by the graphs of the functions:

$$
y=(x+1)^{2}, y^{2}=x+1
$$
## Solution

Integrals 14-11

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-21.jpg?height=1559&width=1428&top_left_y=1039&top_left_x=445)

$$
y=(x+1)^{2}, y^{2}=x+1
$$

The function $y^{2}=x+1$ can be represented as:

$$
y= \pm \sqrt{x+1}
$$

Since $y=(x+1)^{2} \geq 0$ for all $x$, we will use the function $y=\sqrt{x+1}$; i.e., the required figure is bounded by the graphs of $y=(x+1)^{2}$ and $y=\sqrt{x+1}$.

Find the limits of integration:

$$
(x+1)^{2}-\sqrt{x+1}=0
$$

Make the substitution $t=\sqrt{x+1}$, then we get:

$$
\begin{aligned}
& t^{4}-t=0 \\
& t \cdot\left(t^{3}-1\right)=0 \\
& t=0 \text { or } t=1
\end{aligned}
$$

Then:

$$
\begin{aligned}
& t=0 \Rightarrow x=-1 \\
& t=1 \Rightarrow x=0
\end{aligned}
$$

Therefore, the limits of integration are $x \in[-1 ; 0]$.

Find the area of the figure as the difference between the areas of $y=\sqrt{x+1}$ and $y=(x+1)^{2}$

$$
\begin{aligned}
& S=\int_{-1}^{0}(\sqrt{x+1}) d x-\int_{-1}^{0}(x+1)^{2} d x= \\
& =\int_{-1}^{0}(x+1)^{\frac{1}{2}} d(x+1)-\int_{-1}^{0}(x+1)^{2} d(x+1)= \\
& =\left.\frac{2}{3}(x+1)^{\frac{3}{2}}\right|_{-1} ^{0}-\left.\frac{1}{3}(x+1)^{3}\right|_{-1} ^{0}= \\
& =\left(\frac{2}{3} \cdot 1-\frac{2}{3} \cdot 0\right)-\left(\frac{1}{3} \cdot 1-\frac{1}{3} \cdot 0\right)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-11$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-12

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