Question

Use midpoints to approxiamte the area under the curve y=f(x) 5 sin(xx)!+ 2.5 cos(4xx) on the interval [0,1] using 10 equal subdivisions.

Answer 1

Subdividing [0, 1] into 10 equally spaced intervals of length
`\Delta x=(1-0)/(10)=\frac1{10}` gives the partition

`[0,1] = \left[0,\frac1{10}\right] \cup \left[\frac1{10},\frac2{10}\right] \cup \cdots \cup \left[\frac9{10},1\right]`

The
`i`-th subinterval has left and right endpoints, respectively, given by

`\ell_i = (i-1)/(10) \text{ and } r_i = \frac i{10}`

where
`i\in\{1,2,3,\ldots,10\}`.

The midpoint of the
`i`-th interval is the average of these,

`m_i = \frac{\ell_i+r_i}2 = (2i-1)/(20) \in \left\{\frac1{20},\frac3{20},\frac5{20},\ldots,(19)/(20)\right\}`

We approximate the area under
`f(x)` over [0, 1] by the Riemann sum,

`\displaystyle \int_0^1 f(x) \, dx \approx \sum_(i=1)^(10) f(m_i) \Delta x \\\\ ~~~~~~~~ = \frac1{10} \sum_(i=1)^(10) \bigg(5\sin(\pi m_i) + 2.5 \cos(4\pi m_i)\bigg) \\\\ ~~~~~~~~ = \frac{\sin\left(\frac\pi{20}\right) + \sin\left((3\pi)/(20)\right) + \cdots + \sin\left((19\pi)/(20)\right)}2 \\\\ ~~~~~~~~~~~~ + \frac{\cos\left(\frac\pi5\right) + \cos\left(\frac{3\pi}5\right) + \cdots + \cos\left(\frac{19\pi}5\right)}4 \\\\ ~~~~~~~~ \approx 3.19623`

(D)