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OLEGan [10]
3 years ago
8

HELP!!! find f'(-5) if f(x)=(-4x^2-2x) (-x^2-9)

Advanced Placement (AP)
1 answer:
polet [3.4K]3 years ago
3 0
If what I got was right, I got 260.
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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 subdi
alisha [4.7K]

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

[0,1] = \left[0,\dfrac1{10}\right] \cup \left[\dfrac1{10},\dfrac2{10}\right] \cup \cdots \cup \left[\dfrac9{10},1\right]

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

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

m_i = \dfrac{\ell_i+r_i}2 = \dfrac{2i-1}{20} \in \left\{\dfrac1{20},\dfrac3{20},\dfrac5{20},\ldots,\dfrac{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(\frac{3\pi}{20}\right) + \cdots + \sin\left(\frac{19\pi}{20}\right)}2 \\\\ ~~~~~~~~~~~~ + \dfrac{\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)

6 0
2 years ago
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