## Task Condition

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

$$
y=x^{2} \cdot \cos x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right)
$$
## Solution

Obviously, when $00
$$

And at points $x=0$ and $x=\frac{\pi}{2}$:

$$
x^{2} \cdot \cos x=0
$$

We calculate the area:

$$
S=\int_{0}^{\pi / 2} x^{2} \cdot \cos x d x=
$$

Using integration by parts:

$$
\begin{aligned}
& u=x^{2}, d v=\cos x d x \\
& d u=2 x d x, v=\sin x
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-57.jpg?height=1017&width=1031&top_left_y=985&top_left_x=909)

We get:

$$
\begin{aligned}
& =\left.x^{2} \cdot \sin x\right|_{0} ^{\pi / 2}+\int_{0}^{\pi / 2} 2 x \cdot \sin x d x= \\
& =\left(\frac{\pi}{2}\right)^{2} \cdot \sin \frac{\pi}{2}-0^{2} \cdot \sin 0+2 \cdot \int_{0}^{\pi / 2} x \cdot \sin x d x=\frac{\pi^{2}}{4}+2 \cdot \int_{0}^{\pi / 2} x \cdot \sin x d x=
\end{aligned}
$$

Using integration by parts again:

$$
\begin{aligned}
& u=x, d v=\sin x d x \\
& d u=d x, v=-\cos x
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\frac{\pi^{2}}{4}+2 \cdot\left(\left.x \cdot(-\cos x)\right|_{0} ^{\pi / 2}+\int_{0}^{\pi / 2}(-\cos x) d x\right)= \\
& =\frac{\pi^{2}}{4}+2 \cdot\left(\frac{\pi}{2} \cdot\left(-\cos \frac{\pi}{2}\right)-0 \cdot(-\cos 0)-\left.\sin x\right|_{0} ^{\pi / 2}\right)= \\
& =\frac{\pi^{2}}{4}+2 \cdot\left(\frac{\pi}{2} \cdot 0-0-\left(\sin \frac{\pi}{2}-\sin 0\right)\right)=\frac{\pi^{2}}{4}+2(0-0-(1-0))=\frac{\pi^{2}}{4}-2
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_14-30»

Categories: Kuznetsov's Integral Problems 14 | Integrals

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- Last edited: 07:54, June 10, 2010.
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## Problem Kuznetsov Integrals 14-31

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