5y = 4x + 20
- 4x + 5y = 20
Split up the integration interval into 4 subintervals:
![\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac%5Cpi8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi8%2C%5Cdfrac%5Cpi4%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi4%2C%5Cdfrac%7B3%5Cpi%7D8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%7B3%5Cpi%7D8%2C%5Cdfrac%5Cpi2%5Cright%5D)
The left and right endpoints of the
-th subinterval, respectively, are
![\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8](https://tex.z-dn.net/?f=%5Cell_i%3D%5Cdfrac%7Bi-1%7D4%5Cleft%28%5Cdfrac%5Cpi2-0%5Cright%29%3D%5Cdfrac%7B%28i-1%29%5Cpi%7D8)
![r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8](https://tex.z-dn.net/?f=r_i%3D%5Cdfrac%20i4%5Cleft%28%5Cdfrac%5Cpi2-0%5Cright%29%3D%5Cdfrac%7Bi%5Cpi%7D8)
for
, and the respective midpoints are
![m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8](https://tex.z-dn.net/?f=m_i%3D%5Cdfrac%7B%5Cell_i%2Br_i%7D2%3D%5Cdfrac%7B%282i-1%29%5Cpi%7D8)
We approximate the (signed) area under the curve over each subinterval by
![T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)](https://tex.z-dn.net/?f=T_i%3D%5Cdfrac%7Bf%28%5Cell_i%29%2Bf%28r_i%29%7D2%28%5Cell_i-r_i%29)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4T_i%5Capprox%5Cboxed%7B3.038078%7D)
We approximate the area for each subinterval by
![M_i=f(m_i)(\ell_i-r_i)](https://tex.z-dn.net/?f=M_i%3Df%28m_i%29%28%5Cell_i-r_i%29)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4M_i%5Capprox%5Cboxed%7B2.981137%7D)
We first interpolate the integrand over each subinterval by a quadratic polynomial
, where
![p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}](https://tex.z-dn.net/?f=p_i%28x%29%3Df%28%5Cell_i%29%5Cdfrac%7B%28x-m_i%29%28x-r_i%29%7D%7B%28%5Cell_i-m_i%29%28%5Cell_i-r_i%29%7D%2Bf%28m%29%5Cdfrac%7B%28x-%5Cell_i%29%28x-r_i%29%7D%7B%28m_i-%5Cell_i%29%28m_i-r_i%29%7D%2Bf%28r_i%29%5Cdfrac%7B%28x-%5Cell_i%29%28x-m_i%29%7D%7B%28r_i-%5Cell_i%29%28r_i-m_i%29%7D)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4%5Cint_%7B%5Cell_i%7D%5E%7Br_i%7Dp_i%28x%29%5C%2C%5Cmathrm%20dx)
It so happens that the integral of
reduces nicely to the form you're probably more familiar with,
![S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))](https://tex.z-dn.net/?f=S_i%3D%5Cdisplaystyle%5Cint_%7B%5Cell_i%7D%5E%7Br_i%7Dp_i%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7Br_i-%5Cell_i%7D6%28f%28%5Cell_i%29%2B4f%28m_i%29%2Bf%28r_i%29%29)
Then the integral is approximately
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4S_i%5Capprox%5Cboxed%7B3.000117%7D)
Compare these to the actual value of the integral, 3. I've included plots of the approximations below.
True
Note that:
![secant \theta = \frac{1}{cos\theta}](https://tex.z-dn.net/?f=secant%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7Bcos%5Ctheta%7D)
The graph of sine and cosine functions are very similar. They only have a shift in the x -axis.
That is, sin x = cos (90 - x)
Since there is a great similarity between the sine and cosine graphs, and the secant graph is an inverse of the cosine graph, the graph of sine can be used to construct the graph of the secant function
Mathematically:
since sec x = 1 / cos x
and, cos x = sin (90 - x)
therefore, sec x = 1 / sin (90 - x)
The graphs of the sine and secant functions are attached to this solution
Learn more here: brainly.com/question/9554579
Answer:
a. 2
b. 6
c. 1
Step-by-step explanation:
The degree is the highest exponent on the variable in an expression
a. 2
b. 6
Both x and y have exponents of 3. To determine the degree, add the exponents together. 3+3=6
c. 1
When no exponent is present on the variable, it is always 1.