## Task Condition

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

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

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-06.jpg?height=769&width=1105&top_left_y=969&top_left_x=818)

$$
\begin{aligned}
& S=S_{0 x} ; 0 \leq x \leq \pi / 2 ; 0 \leq y \leq \sin x \cos ^{2} x \\
& S=\int_{0}^{\pi / 2} y d x= \\
& =\int_{0}^{\pi / 2}\left(\sin x \cos ^{2} x-0\right) d x= \\
& =-\int_{0}^{\pi / 2} \cos ^{2} x d(\cos x)= \\
& =-\left.\frac{\cos ^{3} x}{3}\right|_{0} ^{\pi / 2}= \\
& =-\left(\frac{\cos ^{3} \pi / 2}{3}-\frac{\cos ^{3} 0}{3}\right)=-\left(\frac{0^{3}}{3}-\frac{1^{3}}{3}\right)=\frac{1}{3}
\end{aligned}
$$

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} \% \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}+14-4 »

Categories: Kuznetsov Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-5

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