## Calculus (3rd Edition)

$V\approx 2.4649$
Given $$y=\cos x ; \quad\left[0, \frac{\pi}{2}\right] ; \quad x \text { -axis; } \quad M_{8}$$ Since $$V= \pi \int_{0}^{\pi/2} [f(x)]^2dx= \pi \int_{0}^{\pi/2} \cos^2xdx$$ Now, we will evaluate the integral using $M_8$, since $\Delta x=\dfrac{b-a}{n}=\dfrac{\pi/2}{8}=\dfrac{\pi}{16}$ Therefore, the sub intervals consist of $$[0,\pi/16],[\pi/16, 2\pi/16],[2\pi/16],\cdots ,[7\pi/16,\pi/2]$$ The midpoints of these sub intervals are $$\{ \pi/32,\ 3\pi/32,\ \ \cdots 15\pi/32 \}$$ Hence \begin{align*} M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\ M_{4}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_{8})\right]\Delta x\\ &=\frac{\pi}{32}\left[ f(\pi/32)+ f(3\pi/32)+\cdots+ f(15\pi/32)\right]\\ &\approx 0.785 \end{align*} Hence, $V\approx 2.4649$