307. $y^{2}=9 x, x=16, x=25$ and $y=0$ (Fig. 150).
Solution. For any $x \in [16,25]$, the function $y=\sqrt{9 x}$ takes positive values; therefore, to calculate the area of the given curvilinear trapezoid, we should use formula (1):

$$
S=\int_{16}^{25} \sqrt{9} x d x=\int_{16}^{25} 3 x^{1 / 2} d x=\left.3 \frac{x^{3 / 2}}{3 / 2}\right|_{16} ^{25}=\left.2 x \sqrt{x}\right|_{16} ^{25}=2(125-64)=2 \cdot 61=
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-321.jpg?height=306&width=373&top_left_y=1508&top_left_x=119)

Fig. 150 $=122$ (sq. units).